Maxima and minima in real-life problems
Write the quantity to optimise in one variable using the given condition, find where its derivative is zero, and confirm with the second derivative test.
The method at a glance
The method, step by step
- 1Read the problem. Draw a figure and name the variables, with units.
- 2Write the quantity Q to be made greatest or least as a formula.
- 3★Use the given condition to write Q in terms of one variable only. Note the values that variable can take.
- 4Find dQ/dx and solve dQ/dx = 0. Reject values outside the allowed range.
- 5Find d²Q/dx² at that value: negative means a maximum, positive means a minimum.
- 6Answer the question asked (a length, the greatest area) with units.
★ key step
Example 1: fencing a field beside a canal
A farmer in Punjab has 200 m of fencing. He wants a rectangular field beside a straight canal, with no fence along the canal. Find the dimensions that give the largest area.
- Let each side at right angles to the canal be x m. The side opposite the canal is then (200 − 2x) m.
- Area A = x(200 − 2x) = 200x − 2x², with 0 < x < 100.
- dA/dx = 200 − 4x = 0 gives x = 50.
- d²A/dx² = −4 < 0, so A is greatest at x = 50.
- The other side = 200 − 2 × 50 = 100 m.
- Greatest area = 50 × 100 = 5000 m².
The field should be 50 m by 100 m, with the 100 m side parallel to the canal. Its area is 5000 m².
Example 2: an open box from a square sheet
A square tin sheet of side 24 cm is made into an open box by cutting equal squares from each corner and folding up the sides. Find the side of the square to be cut off for the largest volume, and that volume.
- Let each square cut off have side x cm. The base is a square of side (24 − 2x) cm and the height is x cm, with 0 < x < 12.
- Volume V = x(24 − 2x)² = 4x³ − 96x² + 576x
- dV/dx = 12x² − 192x + 576 = 12(x² − 16x + 48) = 12(x − 4)(x − 12)
- dV/dx = 0 gives x = 4 or x = 12. Reject x = 12, as the base would vanish.
- d²V/dx² = 24x − 192. At x = 4 it is 96 − 192 = −96 < 0, so V is greatest.
- V = 4 × (24 − 8)² = 4 × 256 = 1024 cm³
Cut squares of side 4 cm. The greatest volume is 1024 cm³.
How to draw the figure
- Draw the shape (rectangle, box, cylinder or cone) neatly in pencil.
- Mark each changing length with a letter, e.g. x and 24 − 2x for the box.
- Write the fixed quantity beside the figure, e.g. 'fence 200 m' or 'V = 32 m³'.
- For a box, draw the flat sheet with the four corner squares shaded, then the folded box.
- For a shape inside a circle or semicircle, draw the radius to a corner and mark the right angle.
Results worth remembering
- Of all rectangles with a fixed perimeter, the square has the greatest area.
- Of all rectangles with a fixed area, the square has the least perimeter.
- A closed cylinder of fixed volume has the least surface area when h = 2r.
- An open box from a square sheet of side a has the greatest volume when squares of side a/6 are cut off.
- Always reject a value that makes a length zero or negative, and say why.
A dairy cooperative ordering milk cans wants each can to hold a fixed volume with the least sheet metal. Making the height equal to the diameter does exactly that, and saves money for every farmer member.
2 marks: “Find two positive numbers whose sum is 16 and whose product is the greatest.” P = x(16 − x) = 16x − x². dP/dx = 16 − 2x = 0 gives x = 8, and d²P/dx² = −2 < 0, so P is greatest. The numbers are 8 and 8.
3 marks: “Find the greatest area of a rectangle whose perimeter is 36 cm.” Sides x and 18 − x, so A = 18x − x². dA/dx = 18 − 2x = 0 gives x = 9, and d²A/dx² = −2 < 0. A square of side 9 cm, greatest area 81 cm².
5 marks: “Show that, of all closed right circular cylinders of a given volume, the one with the least total surface area has its height equal to its diameter.” V = πr²h is fixed, so h = V/(πr²). S = 2πr² + 2πrh = 2πr² + 2V/r. dS/dr = 4πr − 2V/r² = 0 gives V = 2πr³, so πr²h = 2πr³ and h = 2r. d²S/dr² = 4π + 4V/r³ > 0, so S is least. Height = 2r = diameter.
Draw it, write it, squeeze it to one variable, then test it.