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Optimisation

Form a function from a practical context, differentiate to find maximum or minimum values, and justify the optimal solution using the second derivative or other tests.

Worked examples

Maximising area with a fixed perimeter

Straightforward

Problem

A farmer uses 80 m of fencing to enclose a rectangular paddock against a river (the river forms one side; no fencing needed there). Find the dimensions that maximise the area.

Minimising material for a container

Moderate

Problem

An open-top cylindrical tin must hold 500π500\pi cm3^3 of paint. Find the radius that minimises the amount of metal used (i.e. minimises the surface area S=πr2+2πrhS = \pi r^2 + 2\pi r h).

Optimisation from a word problem (box)

Challenging

Problem

A 20 cm ×\times 20 cm square sheet of cardboard has equal squares of side xx cm cut from each corner. The sides are folded up to form an open box. Find the value of xx that maximises the volume.

Practise

Q1·Straightforward
A rectangle has a perimeter of 40 cm. If its length is xx cm, its width is (20−x)(20 - x) cm. What value of xx maximises the area?
Q2·Straightforward
The daily profit (in dollars) from selling xx items is P(x)=−2x2+80x−200P(x) = -2x^2 + 80x - 200. How many items per day maximise profit?
Q3·Straightforward
A ball is thrown so that its height above ground (in metres) is h(t)=20t−5t2h(t) = 20t - 5t^2 after tt seconds. What is the maximum height reached, in metres?
Q4·Straightforward
The number of bacteria in a culture after tt hours is N(t)=−t2+8t+20N(t) = -t^2 + 8t + 20 for 0≤t≤100 \leq t \leq 10. At what time (in hours) is the population at its maximum?
Q5·Moderate
A farmer has 120 m of fencing to enclose a rectangular paddock against a straight wall (the wall forms one side and needs no fencing). What length (in metres) should each side perpendicular to the wall be, to maximise the enclosed area?
Q6·Moderate
A piece of wire 60 cm long is bent into a rectangle. What is the maximum possible area, in cm²?
Q7·Moderate
Two positive numbers sum to 12. Find the maximum value of their product.
Q8·Moderate
A closed cylindrical can (open top) has volume 125π125\pi cm³. Its surface area is S=πr2+2πrhS = \pi r^2 + 2\pi r h. Using the constraint πr2h=125π\pi r^2 h = 125\pi (so h=125r2h = \dfrac{125}{r^2}), find the radius rr (in cm) that minimises the surface area.
Q9·Moderate
The revenue from selling xx units is R(x)=500x−2x2R(x) = 500x - 2x^2 dollars and the cost is C(x)=100x+3000C(x) = 100x + 3000 dollars. Find the number of units xx that maximises profit.
Q10·Challenging
A rectangular piece of cardboard measures 16 cm × 10 cm. Equal squares of side xx cm are cut from each corner, and the sides are folded up to make an open box. Find the value of xx (in cm) that maximises the volume.
Q11·Challenging
A closed rectangular box has a square base of side xx cm and height hh cm. Its total surface area is 600600 cm². Given h=600−2x24xh = \dfrac{600 - 2x^2}{4x}, find the value of xx (in cm) that maximises the volume V=x2hV = x^2 h.
Q12·Challenging
A rectangle is inscribed in a semicircle of radius 4. The base of the rectangle lies along the diameter. If half the base has length xx, the height is 16−x2\sqrt{16 - x^2}. Find the maximum area of the rectangle.