Question
A shop uses the following model to estimate n , the number of smoothies sold per day, in terms of x , the price of a single smoothie in pesos.
$$n = \frac{40000}{x^2}$$
The maximum number of smoothies the shop can make in a day is 400.
(a) Find the maximum price they could charge per smoothie for the shop to sell 400 in one day.
(b) On a day when the shop sells smoothies at 50 pesos each, use the model to find
(i) the number of smoothies sold.
(ii) the total income from the smoothies sold.
The cost of making each smoothie is 20 pesos. The profit per day (P) is the total income from the sale of smoothies that day minus the cost of making them.
(c) (i) Show that, according to the model,
$$P= \frac{40000}{x} – \frac{800000}{x^2}$$
(ii) Find $\frac{dP}{dx}$
(iii) Find the value of x for which $\frac{dP}{dx} = 0$
(iv) Find the number of smoothies sold when the profit is maximized.
▶️Answer/Explanation
Detail Solution: –
Let’s dive into this smoothie shop problem step by step, using the given model \( n = \frac{40000}{x^2} \), where \( n \) is the number of smoothies sold per day and \( x \) is the price per smoothie in pesos. The shop can make a maximum of 400 smoothies per day, and we’ll work through each part systematically.
Part(a) Maximum price to sell 400 smoothies in one day
We need to find the price \( x \) such that the number of smoothies sold, \( n \), equals the maximum capacity of 400. Using the model:
\[ n = \frac{40000}{x^2} \]
Set \( n = 400 \):
\[ 400 = \frac{40000}{x^2} \]
Solve for \( x^2 \):
\[ x^2 = \frac{40000}{400} = 100 \]
\[ x = \sqrt{100} = 10 \]
Since \( x \) represents price in pesos, and negative prices don’t make sense, we take the positive root:
\[ x = 10 \]
So, the maximum price they could charge to sell 400 smoothies is 10 pesos.
part (b) At 50 pesos per smoothie
(i) Number of smoothies sold
Now, the price is \( x = 50 \) pesos. Plug this into the model:
\[ n = \frac{40000}{x^2} = \frac{40000}{50^2} = \frac{40000}{2500} = 16 \]
The shop sells 16 smoothies. This is less than 400, which makes sense—the higher the price, the fewer sold, and 16 is within the shop’s capacity.
(ii) Total income
Total income is the number of smoothies sold times the price per smoothie:
\[ \text{Income} = n \cdot x = 16 \cdot 50 = 800 \]
The total income is 800 pesos.
Part(c) Profit-related questions
The profit \( P \) is defined as total income minus total cost, with a cost of 20 pesos per smoothie.
(i) Show that \( P = \frac{40000}{x} – \frac{800000}{x^2} \)
In come: Revenue is price times quantity sold, so:
\[ \text{Income} = x \cdot n = x \cdot \frac{40000}{x^2} = \frac{40000}{x} \]
Cost: Cost is 20 pesos per smoothie, times the number sold:
\[ \text{Cost} = 20 \cdot n = 20 \cdot \frac{40000}{x^2} = \frac{800000}{x^2} \]
Profit: \( P = \text{Income} – \text{Cost} \):
\[ P = \frac{40000}{x} – \frac{800000}{x^2} \]
This matches the given expression, so it’s verified.
(ii) Find \( \frac{dP}{dx} \)
Differentiate \( P = \frac{40000}{x} – \frac{800000}{x^2} \) with respect to \( x \). Rewrite using exponents:
\[ P = 40000 x^{-1} – 800000 x^{-2} \]
Derivative of \( 40000 x^{-1} \):
\[ \frac{d}{dx} (40000 x^{-1}) = 40000 \cdot (-1) x^{-2} = -\frac{40000}{x^2} \]
Derivative of \( -800000 x^{-2} \):
\[ \frac{d}{dx} (-800000 x^{-2}) = -800000 \cdot (-2) x^{-3} = \frac{1600000}{x^3} \]
So:
\[ \frac{dP}{dx} = -\frac{40000}{x^2} + \frac{1600000}{x^3} \]
(iii) Find \( x \) where \( \frac{dP}{dx} = 0 \)
Set the derivative equal to zero to find the critical point:
\[ -\frac{40000}{x^2} + \frac{1600000}{x^3} = 0 \]
\[ \frac{1600000}{x^3} = \frac{40000}{x^2} \]
Multiply through by \( x^3 \) (since \( x \neq 0 \)):
\[ 1600000 = 40000 x \]
\[ x = \frac{1600000}{40000} = 40 \]
So, \( x = 40 \) pesos is where the derivative is zero.
(iv) Number of smoothies sold when profit is maximized
Using \( x = 40 \) in the model:
\[ n = \frac{40000}{x^2} = \frac{40000}{40^2} = \frac{40000}{1600} = 25 \]
The shop sells 25 smoothies when profit is maximized. This is below 400, which is consistent with the capacity constraint not being a limit here.
————Markscheme—————–
(a) $\frac{40000}{x^{2}} = 400$
x = 10 (pesos) (since x is positive)
(b) (i) $\left(\frac{40000}{50^{2}}\right) = 16$
(ii) (50×16) = 800 (pesos)
(c) (i) EITHER
profit for each smoothie = x – 20
$P = \frac{40000}{x^{2}} \times (x – 20)$
OR
profit = revenue – costs = nx – 20n
$P = x \times \frac{40000}{x^{2}} – 20 \times \frac{40000}{x^{2}}$
THEN
$P = \frac{40000}{x} – \frac{800000}{x^{2}}$
(ii) attempt to express P ready for power rule
$P = 40000x^{-1} – 800000x^{-2}$
$\frac{dP}{dx} = -\frac{40000}{x^{2}} + \frac{1600000}{x^{3}}$
OR $\frac{dP}{dx} = -40000x^{-2} + 1600000x^{-3}$
(iii) attempt to find x-value
e.g. sketch of $\frac{dP}{dx}$ with x-intercept indicated OR recognition that it occurs maximum of P OR algebraic approach (requires multiplication by $x^{3}$)
$x=40$
(iv) attempt to substitue their x-value into equation for n
$n=\frac{40000}{40^{2}}$
= 25
Question
The company Fred Express delivers packages. From past experience, the time taken, T , to deliver a package follows a normal distribution with mean 64 hours and standard deviation 12 hours.
(a) State P (T < 64) .
(b) Find P (44 < T < 64) .
30 % of packages are delivered in less than k hours.
(c) (i) Sketch a diagram of this normal distribution, shading the region that represents P (T < k ) .
(ii) Find the value of k .
For quality control, the manager randomly selects five outgoing packages. These selections are independent.
(d) Find the probability that exactly two of these packages are delivered in less than k hours.
Fred Express charges a fixed amount of $4.50 for any package weighing 1 kg or less.
Heavier packages are charged an additional fee of $2.00 per kg. This fee is applied for any weight in excess of 1 kg. For example, a 1.5 kg package is charged an additional $1.00.
(e) Write down an expression for the amount charged to deliver a package of weight x kg, where x > 1 .
(f) Find the amount Fred Express charges for a 5.3 kg package.
Meiling is charged $7.20 for the delivery of a package.
(g) Find the weight of Meiling’s package.
▶️Answer/Explanation
Detail Solution:
Let’s dive into this problem step by step, solving each part with clarity and precision. We’re dealing with a normal distribution for delivery times and a pricing structure for packages, so we’ll tackle the probability questions first and then move to the cost-related ones.
Part(a) State P(T < 64):
The delivery time \( T \) follows a normal distribution with a mean of 64 hours and a standard deviation of 12 hours. In a normal distribution, the mean splits the probability evenly: 50% of the data lies below the mean, and 50% lies above it. Since 64 is the mean, the probability that \( T \) is less than 64 is simply:
\[ P(T < 64) = 0.5 \]
part(b) Find P(44 < T < 64):
We need the probability that the delivery time is between 44 and 64 hours. To compute this, we standardize the values using the z-score formula:
\[ z = \frac{T – \mu}{\sigma} \]
where \( \mu = 64 \) and \( \sigma = 12 \).
For \( T = 44 \):
\[ z = \frac{44 – 64}{12} = \frac{-20}{12} = -1.6667 \]
For \( T = 64 \):
\[ z = \frac{64 – 64}{12} = 0 \]
We want \( P(44 < T < 64) = P(-1.6667 < Z < 0) \), where \( Z \) is the standard normal variable. Using standard normal tables (or approximating):
\( P(Z < 0) = 0.5 \)
\( P(Z < -1.6667) \approx P(Z < -1.67) \approx 0.0475 \) (from standard tables, \( P(Z < 1.67) = 0.9525 \), so \( P(Z < -1.67) = 1 – 0.9525 = 0.0475 \)).
Thus:
\[ P(-1.6667 < Z < 0) = P(Z < 0) – P(Z < -1.6667) = 0.5 – 0.0475 = 0.4525 \]
So, \( P(44 < T < 64) = 0.4525 \).
Part(c) (i) Sketch a diagram of this normal distribution, shading the region that represents P(T < k):
Imagine a bell curve centered at 64 hours. The x-axis represents time \( T \), with 64 at the peak. Mark a point \( k \) to the left of 64 (since 30% of packages are delivered in less than \( k \) hours, \( k < 64 \)). Shade the area under the curve from the left tail up to \( k \). This shaded region represents \( P(T < k) = 0.3 \). I won’t draw it here, but visualize a left-tailed area under the normal curve.
Part(c) (ii) Find the value of k:
We’re told 30% of packages are delivered in less than \( k \) hours, so \( P(T < k) = 0.3 \). Convert this to a z-score:
For \( P(Z < z) = 0.3 \), we look up the standard normal table. The z-score where the cumulative probability is 0.3 is approximately \( z = -0.524 \) (since \( P(Z < -0.524) \approx 0.3 \)).
Now, use the z-score formula to find \( k \):
\[ z = \frac{k – \mu}{\sigma} \]
\[ -0.524 = \frac{k – 64}{12} \]
\[ k – 64 = -0.524 \times 12 = -6.288 \]
\[ k = 64 – 6.288 = 57.712 \]
So, \( k \approx 57.71 \) hours (rounded to 2 decimal places).
Part(d) Find the probability that exactly two of five packages are delivered in less than k hours:
From part (c), \( P(T < k) = 0.3 \), so the probability a package is delivered in less than \( k \) hours is 0.3, and the probability it’s not is \( 1 – 0.3 = 0.7 \). The selections are independent, and we have 5 packages. This is a binomial probability problem:
Number of trials, \( n = 5 \)
Success probability, \( p = 0.3 \)
Probability of exactly 2 successes: \( P(X = 2) = \binom{5}{2} (0.3)^2 (0.7)^3 \)
Calculate:
\( \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10 \)
\( (0.3)^2 = 0.09 \)
\( (0.7)^3 = 0.7 \times 0.7 \times 0.7 = 0.343 \)
\( P(X = 2) = 10 \times 0.09 \times 0.343 = 10 \times 0.03087 = 0.3087 \)
So, the probability is \( 0.3087 \).
Part(e) Write down an expression for the amount charged to deliver a package of weight x kg, where x > 1:
The pricing rule is:
\($4.50\) flat rate for packages up to 1 kg.
For weight above 1 kg, add \( $2.00\) per kg of excess weight.
For \( x > 1 \), the excess weight is \( x – 1 \), and the additional fee is \( 2 \times (x – 1) \). Total cost:
\[ C(x) = 4.50 + 2(x – 1) \]
Simplify:
\[ C(x) = 4.50 + 2x – 2 = 2x + 2.50 \]
So, the expression is \( C(x) = 2x + 2.50 \) dollars, for \( x > 1 \).
Part(f) Find the amount Fred Express charges for a 5.3 kg package:
Since \( 5.3 > 1 \), use the expression from (e):
\[ C(5.3) = 2 \times 5.3 + 2.50 = 10.6 + 2.50 = 13.10 \]
The charge is $13.10.
Part(g) Find the weight of Meiling’s package if she’s charged $7.20:
Assume the package weighs$ \( x > 1 \) kg (since $7.20 > $4.50, the cost for$ \( x \leq 1 \)).$ Use the cost expression:
\[ 7.20 = 2x + 2.50 \]
\[ 7.20 – 2.50 = 2x \]
\[ 4.70 = 2x \]
\[ x = \frac{4.70}{2} = 2.35 \]
The weight is 2.35 kg. (Check: \( C(2.35) = 2 \times 2.35 + 2.50 = 4.70 + 2.50 = 7.20 \), which matches.)
————Markscheme—————–
a) 0.5
(b) 0.452 (0.452209…)
(c) (i)
(ii) 
(ii) P (T<k) = 0.3
Solving a cumulative distribution function OR use of inverse function on GDC k = 57.7 (57.7071…)
(d) recognizing binomial distribution B(5, 0.3) (P(X = 2))
0.309 (0.3087)
(e) 2(x-1) + 4.5 OR 2x + 2.5
(f) $13.10
(g) attempt to solve 2(x-1)+4.5=7.2 OR 2x+2.5=7.2
$2.35 (kg)$
Question 3. [Maximum mark: 15]
A hollow chocolate box is manufactured in the form of a right prism with a regular hexagonal base. The height of the prism is h cm, and the top and base of the prism have sides of length x cm.
a. Given that \(sin 60\frac{\sqrt{3}}{2}\) show that the area of the base of the box is equal to \(\frac{3\sqrt{3x^{2}}}{2}\) [2]
b. Given that the total external surface area of the box is 1200 cm2, show that the volume of the box may be expressed as V = 300 \(\sqrt{3x}-\frac{9}{4}x^{3}\) [5]
c. Sketch the graph of V = 300 \(\sqrt{3x}-\frac{9}{4}x^{3}\) for 0 ≤ x ≤ 16. [2]
d. Find an expression for \(\frac{dV}{dx}\) . [2]
e. Find the value of x which maximizes the volume of the box. [2]
f. Hence, or otherwise, find the maximum possible volume of the box. [2]
▶️Answer/Explanation
(a) splitting diagram into equilateral triangles area \(= 6(\frac{1}{2}x^{2}sin60)\)
\(=\frac{3\sqrt{3}x^{2}}{2}\)
(b) total surface area of prism
\(1200= 2 (3x^{2}\frac{\sqrt{3}}{2})+6xh\)
\(h= \frac{400-\sqrt{3x^{2}}}{2x}\)
\(volume of prism =\frac{3\sqrt{3}}{2}x^{2}h\)
\(= \frac{3\sqrt{3}}{2}x^{2}(\frac{400-\sqrt{3x^{2}}}{2x})
\(= 300\sqrt{3x}-\frac{9}{4}x^{3}\)
(c) 
(d)\(\frac{dV}{dx}= 300\sqrt{3}-\frac{27}{4}x^{2}\)
(e) from the graph of V or\(\frac{dV}{dx}OR solving \frac{dV}{dx}= 0\) \( x = 8.77 (8.77382…)\)
(f) from the graph of V OR substituting their value for x into V
\(V_{max}= 3040 cm^{2}\rightarrow (3039.34…)\)
Question 4. [Maximum mark: 14]
Charlotte decides to model the shape of a cupcake to calculate its volume.
From rotating a photograph of her cupcake she estimates that its cross-section passes through the points (0 , 3.5), (4 , 6), (6.5 , 4), (7 , 3) and (7.5 , 0), where all units are in centimetres. The cross-section is symmetrical in the x-axis, as shown below:
She models the section from (0 , 3.5) to (4 , 6) as a straight line.
a. Find the equation of the line passing through these two points. [2]
Charlotte models the section of the cupcake that passes through the points (4 , 6), (6.5 , 4), (7 , 3) and (7.5 , 0) with a quadratic curve.
b.(i) Find the equation of the least squares regression quadratic curve for these four points.
b. (ii) By considering the gradient of this curve when x = 4 , explain why it may not be a good model. [3]
Charlotte thinks that a quadratic with a maximum point at (4 , 6) and that passes through the point (7.5 , 0) would be a better fit.
c. Find the equation of the new model. [4]
Believing this to be a better model for her cupcake, Charlotte finds the volume of revolution about the x-axis to estimate the volume of the cupcake.
d. (i) Write down an expression for her estimate of the volume as a sum of two integrals.
d. (ii) Find the value of Charlotte’s estimate. [5]
▶️Answer/Explanation
(a) \(y = \frac{5}{8}x+ \frac{7}{2}(y = 0.625x+ 3.5)\)
(b)(i) \(y= -0.975x^{2}+9.5x-16.7\)
\(y= -0974630x^{2}+9.55919x- 16.6569..)\)
(ii) gradient of curve is positive at x = 4
(c) let \(y=ax^{2}+bx+c\) differentiating or using \(x=\frac{-b}{2a}8a+b=0\) substituting in the coordinates
\(7.5^{2}a+ 7.5b+c=0
\(4^{2}a+ 4b+ c=6\)
solve to get \( y =-\frac{24}{49}x^{2}+\frac{192}{49}x-\frac{90}{49}\) OR
\(y = -0.490x^{2}+3.92x-1.84\)
(d)(i) \(\pi \int ^{4}_{0}(\frac{5}{8}x+3.5)^{2}dx+\pi \int ^{7.5}_{4}(-\frac{24}{49}(x-4^{2})+6)^{2}dx\)
(ii) 501cm3(501.189…)
Question
Given \(f (x) = x^2 − 3x^{−1}, x \in {\mathbb{R}}, – 5 \leqslant x \leqslant 5, x \ne 0\),
A football is kicked from a point A (a, 0), 0 < a < 10 on the ground towards a goal to the right of A.
The ball follows a path that can be modelled by part of the graph
\(y = − 0.021x^2 + 1.245x − 6.01, x \in {\mathbb{R}}, y \geqslant 0\).
x is the horizontal distance of the ball from the origin
y is the height above the ground
Both x and y are measured in metres.
i.a.Write down the equation of the vertical asymptote.[1]
i.b.Find \(f ′(x)\).[2]
i.c.Using your graphic display calculator or otherwise, write down the coordinates of any point where the graph of \(y = f (x)\) has zero gradient.[2]
i.d.Write down all intervals in the given domain for which \(f (x)\) is increasing.[3]
ii.a.Using your graphic display calculator or otherwise, find the value of a.[1]
ii.b.Find \(\frac{{dy}}{{dx}}\).[2]
ii.c.(i) Use your answer to part (b) to calculate the horizontal distance the ball has travelled from A when its height is a maximum.
(ii) Find the maximum vertical height reached by the football.[4]
ii.d.Draw a graph showing the path of the football from the point where it is kicked to the point where it hits the ground again. Use 1 cm to represent 5 m on the horizontal axis and 1 cm to represent 2 m on the vertical scale.[4]
ii.e.The goal posts are 35 m from the point where the ball is kicked.
At what height does the ball pass over the goal posts?[2]
▶️Answer/Explanation
Markscheme
equation of asymptote is x = 0 (A1)
(Must be an equation.)[1 mark]
\(f ‘(x) = 2x + 3x^{-2}\) (or equivalent) (A1) for each term (A1)(A1)[2 marks]
stationary point (–1.14, 3.93) (G1)(G1)(ft)
(-1,4) or similar error is awarded (G0)(G1)(ft). Here and also as follow through in part (d) accept exact values \( – {\left( {\frac{3}{2}} \right)^{\frac{1}{3}}}\)for the x coordinate and \(3{\left( {\frac{3}{2}} \right)^{\frac{2}{3}}}\) for the y coordinate.
OR \(2x + \frac{3}{{{x^2}}} = 0\) or equivalent
Correct coordinates as above (M1)
Follow through from candidate’s \(f ′(x)\). (A1)(ft)[2 marks]
In all alternative answers for (d), follow through from candidate’s x coordinate in part (c).
Alternative answers include:
–1.14 ≤ x < 0, 0 < x < 5 (A1)(A1)(ft)(A1)
OR [–1.14,0), (0,5)
Accept alternative bracket notation for open interval ] [. (Union of these sets is not correct, award (A2) if all else is right in this case.)
OR \( – 1.14 \leqslant x < 5,x \ne 0\)
In all versions 0 must be excluded (A1). -1.14 must be the left bound . 5 must be the right bound (A1). For \(x \geqslant – 1.14\) or \(x > – 1.14\) alone, award (A1). For \( – 1.4 \leqslant x < 0\) together with \(x > 0\) award (A2).[3 marks]
a = 5.30 (3sf) (Allow (5.30, 0) but 5.3 receives an (AP).) (A1)[1 mark]
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 0.042x + 1.245\) (A1) for each term. (A1)(A1)[2 marks]
Unit penalty (UP) is applicable where indicated in the left hand column.
(i) Maximum value when \(f ‘ (x) = 0\), \( – 0.042x + 1.245 = 0\), (M1)
(M1) is for either of the above but at least one must be seen.
(x = 29.6.)
Football has travelled 29.6 – 5.30 = 24.3 m (3sf) horizontally. (A1)(ft)
For answer of 24.3 m with no working or for correct subtraction of 5.3 from candidate’s x-coordinate at the maximum (if not 29.6), award (A1)(d).(UP) (ii) Maximum vertical height, f (29.6) = 12.4 m (M1)(A1)(ft)(G2)
(M1) is for substitution into f of a value seen in part (c)(i). f(24.3) with or without evaluation is awarded (M1)(A0). For any other value without working, award (G0). If lines are seen on the graph in part (d) award (M1) and then (A1) for candidate’s value \( \pm 0.5\) (3sf not required.)[4 marks]
(not to scale)
(A1)(A1)(A1)(ft)(A1)(ft)
Award (A1) for labels (units not required) and scale, (A1)(ft) for max(29.6,12.4), (A1)(ft) for x-intercepts at 5.30 and 53.9, (all coordinates can be within 0.5), (A1) for well-drawn parabola ending at the x-intercepts.[4 marks]
Unit penalty (UP) is applicable where indicated in the left hand column.
(UP) f (40.3) = 10.1 m (3sf).
Follow through from (a). If graph used, award (M1) for lines drawn and (A1) for candidate’s value \( \pm 0.5\). (3sf not required). (M1)(A1)(ft)(G2)[2 marks]
Question
The diagram below shows the graph of a line \(L\) passing through (1, 1) and (2 , 3) and the graph \(P\) of the function \(f (x) = x^2 − 3x − 4\)
a.Find the gradient of the line L.[2]
b.Differentiate \(f (x)\) .[2]
c.Find the coordinates of the point where the tangent to P is parallel to the line L.[3]
d.Find the coordinates of the point where the tangent to P is perpendicular to the line L.[4]
e.Find
(i) the gradient of the tangent to P at the point with coordinates (2, − 6).
(ii) the equation of the tangent to P at this point.[3]
f.State the equation of the axis of symmetry of P.[1]
g.Find the coordinates of the vertex of P and state the gradient of the curve at this point.[3]
▶️Answer/Explanation
Markscheme
for attempt at substituted \(\frac{{ydistance}}{{xdistance}}\) (M1)
gradient = 2 (A1)(G2)[2 marks]
\(2x – 3\) (A1)(A1)
(A1) for \(2x\) , (A1) for \(-3\)[2 marks]
for their \(2x – 3 =\) their gradient and attempt to solve (M1)
\(x = 2.5\) (A1)(ft)
\(y = -5.25\) ((ft) from their x value) (A1)(ft)(G2)[3 marks]
for seeing \(\frac{{ – 1}}{{their(a)}}\) (M1)
solving \(2x – 3 = – \frac{1}{2}\) (or their value) (M1)
x = 1.25 (A1)(ft)(G1)
y = – 6.1875 (A1)(ft)(G1)[4 marks]
(i) \(2 \times 2 – 3 = 1\) ((ft) from (b)) (A1)(ft)(G1)
(ii) \(y = mx + c\) or equivalent method to find \(c \Rightarrow -6 = 2 + c\) (M1)
\(y = x – 8\) (A1)(ft)(G2)[3 marks]
\(x = 1.5\) (A1)[1 mark]
for substituting their answer to part (f) into the equation of the parabola (1.5, −6.25) accept x = 1.5, y = −6.25 (M1)(A1)(ft)(G2)
gradient is zero (accept \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)) (A1)[3 marks]
Question
A closed rectangular box has a height \(y{\text{ cm}}\) and width \(x{\text{ cm}}\). Its length is twice its width. It has a fixed outer surface area of \(300{\text{ c}}{{\text{m}}^2}\) .
i.a.Factorise \(3{x^2} + 13x – 10\).[2]
i.b.Solve the equation \(3{x^2} + 13x – 10 = 0\).[2]
i.c.Consider a function \(f(x) = 3{x^2} + 13x – 10\) .
Find the equation of the axis of symmetry on the graph of this function.[2]
i.d.Consider a function \(f(x) = 3{x^2} + 13x – 10\) .
Calculate the minimum value of this function.[2]
ii.a.Show that \(4{x^2} + 6xy = 300\).[2]
ii.b.Find an expression for \(y\) in terms of \(x\).[2]
ii.c.Hence show that the volume \(V\) of the box is given by \(V = 100x – \frac{4}{3}{x^3}\).[2]
ii.d.Find \(\frac{{{\text{d}}V}}{{{\text{d}}x}}\).[2]
ii.e.(i) Hence find the value of \(x\) and of \(y\) required to make the volume of the box a maximum.
(ii) Calculate the maximum volume.[5]
▶️Answer/Explanation
Markscheme
\((3x – 2)(x + 5)\) (A1)(A1)[2 marks]
\((3x – 2)(x + 5) = 0\)
\(x = \frac{2}{3}\) or \(x = – 5\) (A1)(ft)(A1)(ft)(G2)[2 marks]
\(x = \frac{{ – 13}}{6}{\text{ }}( – 2.17)\) (A1)(A1)(ft)(G2)
Note: (A1) is for \(x = \), (A1) for value. (ft) if value is half way between roots in (b).[2 marks]
Minimum \(y = 3{\left( {\frac{{ – 13}}{6}} \right)^2} + 13\left( {\frac{{ – 13}}{6}} \right) – 10\) (M1)
Note: (M1) for substituting their value of \(x\) from (c) into \(f(x)\) .
\( = – 24.1\) (A1)(ft)(G2)[2 marks]
\({\text{Area}} = 2(2x)x + 2xy + 2(2x)y\) (M1)(A1)
Note: (M1) for using the correct surface area formula (which can be implied if numbers in the correct place). (A1) for using correct numbers.
\(300 = 4{x^2} + 6xy\) (AG)
Note: Final line must be seen or previous (A1) mark is lost.[2 marks]
\(6xy = 300 – 4{x^2}\) (M1)
\(y = \frac{{300 – 4{x^2}}}{{6x}}\) or \(\frac{{150 – 2{x^2}}}{{3x}}\) (A1)[2 marks]
\({\text{Volume}} = x(2x)y\) (M1)
\(V = 2{x^2}\left( {\frac{{300 – 4{x^2}}}{{6x}}} \right)\) (A1)(ft)
\( = 100x – \frac{4}{3}{x^3}\) (AG)
Note: Final line must be seen or previous (A1) mark is lost.[2 marks]
\(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 100 – \frac{{12{x^2}}}{3}\) or \(100 – 4{x^2}\) (A1)(A1)
Note: (A1) for each term.[2 marks]
Unit penalty (UP) is applicable where indicated in the left hand column
(i) For maximum \(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0\) or \(100 – 4{x^2} = 0\) (M1)
\(x = 5\) (A1)(ft)
\(y = \frac{{300 – 4{{(5)}^2}}}{{6(5)}}\) or \(\left( {\frac{{150 – 2{{(5)}^2}}}{{3(5)}}} \right)\) (M1)
\( = \frac{{20}}{3}\) (A1)(ft)
(UP) (ii) \(333\frac{1}{3}{\text{ c}}{{\text{m}}^3}{\text{ }}(333{\text{ c}}{{\text{m}}^3})\)
Note: (ft) from their (e)(i) if working for volume is seen.[5 marks]