Find the least area of sheet metal required to make an open baking dish of square base and vertical sides of capacity 2048cm3

Question

I have to use differentiation

Answer 1

Let surface area be y cm² and the side of the square base be x cm. Let the vertical side be length h cm.

y=x²+4xh and volume=2048=x²h, so h=2048/x².

y=x²+8192/x.

dy/dx=2x-8192/x²=0 at an extremum.

d²y/dx²=2+16384/x³.

2x³=8192, x³=4096, x=16cm, so d²y/dx²>0 (minimum).

y=256+512=768cm² is the minimum surface area, and the least amount of sheet metal required to make the baking dish.

Answer 2

Let the side of the square base be x, and height of the dish be h.

Then Volume, V = x^2 * h  = 2048 cm^3

h = 2048/x^2  -----------------------------(1)

Also, Surface area of the box, A = x^2 + 4xh  ------------------(2)

On putting the value of (1) in (2) we get:

A = x^2 + 4x * 2048/x^2

or A = x^2 + 8192/x

We want to minimize A so,

A' = 2x - 8192/x^2 = 0

or A' = (2x^3 - 8192)/x^2

To construct a box, x cannot be equal to 0 or negative, so:

2x^3 - 8192 = 0

or x^3 = 8192/2 = 4096

or x =  16   (Critical Number)

To verify that x= 16 minimizes A, we use 2nd derivative test.

A'' = 2 + 16384/x^3, which is greater than 0 at x=40, so A is minimum at x=40

h= 2048/x^2

h = 2048/(16^2) =8

Therefore, the least area should be

A = x^2 + 4xh = 16^2 + 4*16*8 = 768 cm^2