A cuboid has a total surface area of 120 cm

^{2}. Its base measures*x*cm by 2*x*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 = 2x

^{2}. 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 = 4x

^{2}+ 6xh120 – 4x

^{2}= 6xhh = (120 – 4x

^{2})/6xh = 20x

^{-1}– 2x/3Given that the volume of the cuboid is

*V*cm^{3},ii) show that

*V*= 40*x*– 4 x^{3}/3 . [1]V = x × 2x × h

V = 2 x

^{2 }(20x^{-1}– 2x/3)V = 40x - 4 x

^{3}/3Given that

*x*can vary,iii)

**show that***V*has a stationary value when*h*= 4x/3 . [4]dV/dx = 40 - 4x

^{2}0 = 40 - 4x

^{2}x = +√10 (reject the negative root)

Substitute this value of x into h = 20x

^{-1}– 2x/3h = 20/√10 - 2√10/3 = 2 √10√10/√10 - 2 √10/3 (Since 20 = 2 × 10 and 10 itself = √10

^{2})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 x

^{3}/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]

d

^{2}V/dx^{2 }= -8x = (negative)(positive) = (negative). Note that x itself must be > 0, i.e., positiveThe function is concave down. The turning point is a maximum.

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