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IB Mathematics AI SL Optimization problems in context MAI Study Notes - New Syllabus

IB Mathematics AI SL Optimization problems in context MAI Study Notes

LEARNING OBJECTIVE

  • Optimisation problems in context

Key Concepts: 

  • Optimisation.

MAI HL and SL Notes – All topics

OPTIMISATION PROBLEMS

Optimization Problems in Context

Optimization involves finding the maximum or minimum values of a function, often subject to constraints, and is widely used in real-life applications such as maximizing profit or minimizing cost.

 Key Steps in Solving Optimization Problems:

  1. Identify the quantity to be maximized or minimized (objective).
  2. Write an equation for that quantity in terms of a variable.
  3. Use any constraints to reduce it to a function of a single variable.
  4. Differentiate to find critical points by solving \( f^\prime(x) = 0 \).
  5. Use second derivative or context to determine if it’s a max/min.

 Common Applications:

  • Maximizing revenue or profit
  • Minimizing cost or material usage
  • Maximizing area or volume given perimeter/surface constraints

Example 

A company finds that the profit \( P \) from selling \( x \) units is given by: \( P(x) = -2x^2 + 80x – 100 \). Find the number of units that should be sold to maximize profit.

▶️Answer/Explanation

Differentiate: \( P'(x) = -4x + 80 \)

Set derivative to 0: \( -4x + 80 = 0 \Rightarrow x = 20 \)

Second derivative: \( P”(x) = -4 \lt 0 \), so maximum.

Answer: Profit is maximized when 20 units are sold.

Example 

A manufacturer finds the cost \( C \) to produce \( x \) items is: \( C(x) = x^2 – 40x + 500 \). Find the number of items that minimizes the cost.

▶️Answer/Explanation

Differentiate: \( C'(x) = 2x – 40 \)

Set derivative to 0: \( 2x – 40 = 0 \Rightarrow x = 20 \)

Second derivative: \( C”(x) = 2 \gt 0 \), so minimum.

Answer: Minimum cost when 20 items are produced.

Example 

Find the dimensions of a closed cylindrical can with a fixed surface area of $600π$ cm² that maximize the volume.

▶️Answer/Explanation

Let radius = \( r \), height = \( h \)

Surface area: \( 2\pi r^2 + 2\pi rh = 600\pi \) ⇒ \( r^2 + rh = 300 \) ⇒ \( h = \frac{300 – r^2}{r} \)

Volume: \( V = \pi r^2 h = \pi r^2 \left( \frac{300 – r^2}{r} \right) = \pi r(300 – r^2) \)

Differentiate: \( V'(r) = \pi(300 – 3r^2) \) Set \( V'(r) = 0 \Rightarrow 300 – 3r^2 = 0 \Rightarrow r = 10 \)

Then \( h = \frac{300 – 100}{10} = 20 \)

Answer: Max volume when radius = 10 cm, height = 20 cm.

Example 

Among all the rectangles of area 25, find the one with the minimum perimeter.

▶️Answer/Explanation

Let \( x \) be one side of the rectangle (this is our variable).

If the other side is \( y \), then:

Area = 25 ⇒ \( xy = 25 \Rightarrow y = \frac{25}{x} \)

The perimeter is: \( P = 2x + 2y = 2x + \frac{50}{x} \)

Differentiate: \( \frac{dP}{dx} = 2 – \frac{50}{x^2} \)

Find stationary points: \( \frac{dP}{dx} = 0 \Rightarrow 2 – \frac{50}{x^2} = 0 \Rightarrow x^2 = 25 \Rightarrow x = 5 \)

Check second derivative: \( \frac{d^2P}{dx^2} = \frac{100}{x^3} \)

At \( x = 5 \), \( \frac{d^2P}{dx^2} > 0 \), so we have a minimum.

Answer: The rectangle with minimum perimeter is a square (when \( x = y = 5 \)), and minimum perimeter is:

\( P_{\text{min}} = 2(5 + 5) = 20 \)

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