Y10 Add Maths Max/min Quiz

A cuboid has a total surface area of 120 cm². Its base measures x cm by 2x cm and its height is h cm.

i) Obtain an expression for h in terms of x.                                                                                                     [2]

Make a sketch:

The area of the base is x × 2x = 2x².  The top face is the same.

The area of the right side is 2x × h = 2xh. The left face is the same.

The area of the front face is x × h = xh.  The back is the same.

Surface area = 120 = 4x² + 6xh

120 – 4x² = 6xh

h = (120 – 4x²)/6x

h = 20x^-1 – 2x/3

Given that the volume of the cuboid is Vcm³,

ii) show that V = 40x – 4 x³/3 .                                                                                                               [1]

V = x × 2x × h

V = 2 x² (20x^-1 – 2x/3)

V = 40x - 4 x³/3

Given that x can vary,

iii) show that V has a stationary value when h = 4x/3 .                                                                            [4]

dV/dx = 40 - 4x²

0 = 40 - 4x²

x = +√10  (reject the negative root)

Substitute this value of x into h = 20x^-1 – 2x/3

h = 20/√10 - 2√10/3 = 2 √10√10/√10 - 2 √10/3    (Since 20 = 2 × 10 and 10 itself = √10²)

h = 2√10 - 2√10/3

h = 6√10/3 - 2√10/3

h = 4√10/3

h = 4x/3      (Recall that x = √10).

iv) Find the stationary value of V.                                                                                                       [1]

V = 40x - 4 x³/3

= 40√10 - 4 √10 ³/3               ( √10) ³  = √10√10√10 = 10√10 )

= 40(√10 - √10)/3)

= 80√10/3

iv) Determine the nature of the stationary value. Show working.                                                          [2]

d²V/dx² = -8x = (negative)(positive) = (negative).    Note that x itself must be > 0, i.e., positive

The function is concave down. The turning point is a maximum.