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Optimisation0%

Differentiation · Topic 12 of 15

Optimisation

Video lesson · from 1:30:002 worked examples

One lesson video covers all of Differentiation, so it opens at 1:30:00 for this topic — not from the beginning.

Theory

Optimisation involves finding the greatest (maximum) or least (minimum) value of a quantity under certain constraints, such as finding the maximum volume of a box or the minimum material needed to make a can.

Steps for Optimisation Problems

Identify the quantity to be maximised or minimised.
Form an equation for this quantity in terms of two variables (e.g. area $A$ in terms of $x$ and $y$ ).
Use the given constraint (e.g. total perimeter) to write one variable in terms of the other.
Substitute this into the main equation to get a function of one variable only (e.g. $A(x)$ ).
Differentiate the function and set the derivative to zero ( $A'(x) = 0$ ) to find stationary points.
Use a nature table (or the second derivative) to confirm whether the point is a maximum or a minimum.
Calculate the maximum/minimum value of the quantity required.

⚠️ Common Examiner Traps

Get to one variable first: use the constraint given in the question to eliminate a variable before differentiating. You cannot differentiate a formula in two unknowns.
Justify the nature: finding the stationary point is not enough — a nature table is needed to show it really is the maximum or minimum.
Check the endpoints too: in a context with a restricted range, the optimum can sit at an end of the interval rather than at a stationary point.
Answer in context, with units: if the question asks for the minimum cost, give the cost, not the value of $x$ that produces it. Read which one is wanted.
Reject impossible solutions: negative lengths, times or prices are not valid answers — say why you are discarding them.

Worked examples

Example 1

A square piece of card of length 30cm has a square of side $x\text{ cm}$ cut from each corner. An open box is formed by turning up the sides.

a) Show that the volume, $V$ , of the box can be expressed by $900x - 120x^2 + 4x^3$ .

b) Find the maximum volume of the box.

Part a) Forming the expression

The base of the box is a square with side length $(30 - 2x)$ .

The height of the box is $x$ .

\begin{aligned} V &= \text{Area of base} \times \text{height} \\ V &= (30 - 2x)^2 \times x \\ V &= (900 - 120x + 4x^2)x \\ V &= 900x - 120x^2 + 4x^3 \end{aligned}

Part b) Maximising Volume

Differentiate and set to 0 for a maximum:

\begin{aligned} V'(x) &= 900 - 240x + 12x^2 \\ 12x^2 - 240x + 900 &= 0 \end{aligned}

Divide by 12:

\begin{aligned} x^2 - 20x + 75 &= 0 \\ (x - 5)(x - 15) &= 0 \end{aligned}

So $x = 5$ or $x = 15$ .

If $x = 15$ , the length of the base would be $30 - 2(15) = 0$ , which is impossible. So x = 5.

Verify it's a maximum using the second derivative factor:

V''(x) = -240 + 24x

At $x = 5$ , $V''(5) = -240 + 24(5) = -120$ . Since $V'' < 0$ , this is a maximum.

Calculate the maximum volume:

\begin{aligned} V(5) &= 900(5) - 120(5)^2 + 4(5)^3 \\ &= 4500 - 120(25) + 4(125) \\ &= 4500 - 3000 + 500 \\ &= 2000 \end{aligned}

The maximum volume of the box is 2000 cm³.

Example 2

An open box with a square base has volume $256\text{ cm}^3$ .

a) Taking $x$ as the length of the base, show that the surface area of the inside of the box can be expressed as $x^2 + \frac{1024}{x}$ .

b) Find the dimensions of the box which will minimize the surface area and calculate its area.

Part a) Forming the expression

Let the dimensions of the base be $x$ by $x$ and the height be $h$ .

The volume is $V = x^2 h = 256$ .

Therefore, $h = \frac{256}{x^2}$ .

The surface area (open box, no top) is:

\begin{aligned} A &= \text{Area of base} + 4 \times \text{Area of sides} \\ A &= x^2 + 4xh \end{aligned}

Substitute $h$ :

\begin{aligned} A &= x^2 + 4x\left(\frac{256}{x^2}\right) \\ &= x^2 + \frac{1024}{x} \end{aligned}

Part b) Minimising Surface Area

A(x) = x^2 + 1024x^{-1}

Differentiate and set to 0 for a minimum:

A'(x) = 2x - 1024x^{-2} = 2x - \frac{1024}{x^2}

\begin{aligned} 2x - \frac{1024}{x^2} &= 0 \\ 2x &= \frac{1024}{x^2} \\ 2x^3 &= 1024 \\ x^3 &= 512 \\ x &= \sqrt[3]{512} = 8 \end{aligned}

Verify it's a minimum using the second derivative:

A''(x) = 2 + 2048x^{-3}

At $x = 8$ , $A''(8) = 2 + \frac{2048}{512} = 2 + 4 = 6$ . Since $A'' > 0$ , this is a minimum.

Dimensions and Minimum Area:

Length and width are $x = 8\text{ cm}$ .

Height $h = \frac{256}{8^2} = \frac{256}{64} = 4\text{ cm}$ .

Dimensions are 8 cm × 8 cm × 4 cm.

Minimum surface area:

\begin{aligned} A(8) &= 8^2 + \frac{1024}{8} \\ &= 64 + 128 \\ &= 192 \end{aligned}

The minimum surface area is 192 cm².