Hello, anthmoo!

Let me walk through this one . . .

A cuboid has volume of 8 m³.

The base of the cuboid is square with sides of length *x* m.

The surface area of the cuboid is *A* m².

(1) Show that: .A .= .2x² + (32/x)

(2) Find: .dA/dx

(3) Find the value of *x* which gives the smallest Surface Area of the cuboid. Code:

x
* - - - *
x/ / |
* - - - * |
| | |y
| | |
y| y| |
| | *
| | /x
* - - - *
x

We are given the volume: .V .= .x²y .= .8 . → . y .= .8/x² . **[1]**

The top and bottom have an area of: 2x²

The four sides have an area of: 4xy

The total surface area is: .A .= .2x² + 4xy . **[2]**

Substitute **[1]** into **[2]**: . A .= .2x² + 4x(8/x²)

Hence, we have: . A .= .2x² + (32/x) . **(1)**

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: .A .= .2x² + 32x^{-1}

Then: .dA/dx .= .4x - 32x^{-2} . **(2)**

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: . 4x - 32x^{-2} .= .0

Multiply by x²: . 4x³ - 32 .= .0

And we have: . 4x³ = 32 . → . x³ = 8 . → . x = 2 .**(3)**

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Substitute into **[1]**: . y .= .8/2² . → . y = 2

As you may have suspected,

. . the cuboid with minimum surface area is a __cube__.