IBDP MAI : Topic 5 Calculus - SL 5.6 Local maximum and minimum points AI HL Paper 2

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