Question

A box, open at the top is to be made from cardboard. The base of the box is a square of side x and its height is y. If the volume of the box is 32u^2, find the dimensions of the box if the area is to be least. Please help?

Answer 1

See the explanation please.

Explanation:

The indicated volume does not make any sense and needs to be
clarified, but for now we can assume the volume to be `32 m^3`:

The area of the base is:

`A_(base)=x^2`

The volume would be:

`V=x^2y=32m^3`

Calculating `y` in terms of `x` in above equation results:

`y=32/x^2`

Total surface area of a box with open top is:

`A=x^2+4xy`

Substituting for `y`:

`A=x^2+4x * 32/x^2`

`A=x^2+128/x`

To find the critical points equate the first derivative to zero:

`(dA)/dx=A'=2x-128/x^2`

`(2x^3 - 128)/x^2=0`

`x^3-64=0`

`x^3=64`

`x=4m`

To verify the nature of the critical point use 2nd derivative test:

`A"=2+256/x^3`

`A"(4)=2+256/4^3=6>0=>` Verifies it is a minimum so:

`x=4m`

`y=32/16=2m`

Thus the box dimensions for a minimum surface area are:

`4m * 4m * 2m`

And the minimum surface area would be:

`A=x^2+4xy=4^2+4*4*2=16+32=48m^2`

Answer 2

`x=4root3[u^2`............ `y=2root3[u^2`

Explanation:

Assuming the units expressed for the volume are correct and are indeed expressed in terms of `u^2` then we may say ,

Volume `V=x^2y=32u^2`........`[1]` , the surface area `A=x^2+4xy`......`[2]`.

From ....`[1]` `y=32u^2/x^2` and substituting this value for `y` in .....`[2]`,

`A=x^2+4x[32u^2]/[x^2]` = `[x^2+[4][32u^2]]/x]`......`[3]`

Differentiating ......`[3]` wrt `x`, `d/dx`= `2x - 4[32u^2]/x^2` =0 [ for max/min and considering `u` constant]

`x^3 = 64u^2`, therefore `x = 4root3[u^2`. substituting this value for `x` in .....`[1]` will give the value of `y` in the answer.

The second derivative =`2+8[32u^2]/x^3` which is positive, since `x` and `u` must be both positive [ must have positive values for a box] and so confirms that the value of `x` obtained will minimise the surface area of the box.