Sec 3 A-Math · Quadratic functions

Maximum and minimum problems questions with answers

Practice questions on maximum and minimum problems, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

A shop's daily profit, $ $P$ , from selling $x$ cakes is $P = 20x - x^2 - 120$ . By completing the square, find the greatest daily profit.

Step 1 of 2

Take out −1, then complete the square

P = -(x^2 - 20x) - 120

= -[(x - 10)^2 - 100] - 120

= -(x - 10)^2 + 100 - 120

= -(x - 10)^2 - 20

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Questions

Question 1From a given formula
A ball is thrown upwards. Its height, $h$ m, $t$ seconds after it is thrown is given by $h = 5 + 10t - 5t^2$ . By completing the square, find the greatest height of the ball.
Show the worked solution
1. Step 1: Take out −5 from the t terms $h = -5(t^2 - 2t) + 5$
2. Step 2: Complete the square inside the bracket, then multiply out the square bracket $= -5[(t - 1)^2 - 1] + 5$ $= -5(t - 1)^2 + 5 + 5$ $= -5(t - 1)^2 + 10$
3. Step 3: The square is never negative, so h is greatest when it is 0 $\text{Greatest height} = 10$ m
Question 2From a given formula
A shop's daily profit, $ $P$ , from selling $x$ cakes is $P = 30x - x^2 - 80$ . By completing the square, find the greatest daily profit.
Show the worked solution
1. Step 1: Take out −1, then complete the square $P = -(x^2 - 30x) - 80$ $= -[(x - 15)^2 - 225] - 80$ $= -(x - 15)^2 + 225 - 80$ $= -(x - 15)^2 + 145$
2. Step 2: The square is never negative, so P is greatest when it is 0 $\text{Greatest profit} = 145$
Question 3Form the formula first
A farmer has $48$ m of fencing to make three sides of a rectangular pen; a long wall forms the fourth side. The two sides at right angles to the wall are each $x$ m long. Find the greatest possible area, in $\text{m}^2$ , of the pen.
Show the worked solution
1. Step 1: The side along the wall uses the rest of the fencing $\text{Length} = 48 - 2x$ $\mathstrut A = x(48 - 2x)$
2. Step 2: Expand, take out the coefficient of x², then complete the square $A = 48x - 2x^2$ $A = -2(x^2 - 24x)$ $= -2[(x - 12)^2 - 144]$ $= -2(x - 12)^2 + 288$
3. Step 3: The square is never negative, so A is greatest when it is 0 $\text{Greatest area} = 288$
Question 4Form the formula first
A rectangle has a perimeter of $20$ cm and a width of $x$ cm. Find the greatest possible area, in $\text{cm}^2$ , of the rectangle.
Show the worked solution
1. Step 1: Length + width is half the perimeter $\text{Length} = 10 - x$ $\mathstrut A = x(10 - x)$
2. Step 2: Expand, take out the coefficient of x², then complete the square $A = 10x - x^2$ $A = -(x^2 - 10x)$ $= -[(x - 5)^2 - 25]$ $= -(x - 5)^2 + 25$
3. Step 3: The square is never negative, so A is greatest when it is 0 $\text{Greatest area} = 25$

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Notes: Quadratic functions Worksheet with answers (PDF)

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Worked example

A shop's daily profit, $

P

, from selling

x

cakes is

P = 20x - x^2 - 120

. By completing the square, find the greatest daily profit.

Step 1 of 2

Take out −1, then complete the square

P = -(x^2 - 20x) - 120

= -[(x - 10)^2 - 100] - 120

= -(x - 10)^2 + 100 - 120

= -(x - 10)^2 - 20

Questions

Question 1From a given formula

A ball is thrown upwards. Its height, $h$ m, $t$ seconds after it is thrown is given by $h = 5 + 10t - 5t^2$ . By completing the square, find the greatest height of the ball.

Show the worked solution

Step 1: Take out −5 from the t terms $h = -5(t^2 - 2t) + 5$
Step 2: Complete the square inside the bracket, then multiply out the square bracket $= -5[(t - 1)^2 - 1] + 5$ $= -5(t - 1)^2 + 5 + 5$ $= -5(t - 1)^2 + 10$
Step 3: The square is never negative, so h is greatest when it is 0 $\text{Greatest height} = 10$ m

Question 2From a given formula

A shop's daily profit, $ $P$ , from selling $x$ cakes is $P = 30x - x^2 - 80$ . By completing the square, find the greatest daily profit.

Show the worked solution

Step 1: Take out −1, then complete the square $P = -(x^2 - 30x) - 80$ $= -[(x - 15)^2 - 225] - 80$ $= -(x - 15)^2 + 225 - 80$ $= -(x - 15)^2 + 145$
Step 2: The square is never negative, so P is greatest when it is 0 $\text{Greatest profit} = 145$

Question 3Form the formula first

A farmer has $48$ m of fencing to make three sides of a rectangular pen; a long wall forms the fourth side. The two sides at right angles to the wall are each $x$ m long. Find the greatest possible area, in $\text{m}^2$ , of the pen.

Show the worked solution

Step 1: The side along the wall uses the rest of the fencing $\text{Length} = 48 - 2x$ $\mathstrut A = x(48 - 2x)$
Step 2: Expand, take out the coefficient of x², then complete the square $A = 48x - 2x^2$ $A = -2(x^2 - 24x)$ $= -2[(x - 12)^2 - 144]$ $= -2(x - 12)^2 + 288$
Step 3: The square is never negative, so A is greatest when it is 0 $\text{Greatest area} = 288$

Question 4Form the formula first

A rectangle has a perimeter of $20$ cm and a width of $x$ cm. Find the greatest possible area, in $\text{cm}^2$ , of the rectangle.

Show the worked solution

Step 1: Length + width is half the perimeter $\text{Length} = 10 - x$ $\mathstrut A = x(10 - x)$
Step 2: Expand, take out the coefficient of x², then complete the square $A = 10x - x^2$ $A = -(x^2 - 10x)$ $= -[(x - 5)^2 - 25]$ $= -(x - 5)^2 + 25$
Step 3: The square is never negative, so A is greatest when it is 0 $\text{Greatest area} = 25$