Question

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h and base length b of the box be so as to maximize its volume

Answer 1

`b=h=√(6)` m

Explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box=
`2b^2+4bh`

`2b^2+4bh=36`

`b^2+2bh=18`

`2bh=18-b^2`

`h=(18-b^2)/(2b)`

Volume of box, V=
`b^2h`

Substitute the values

`V=b^2* (18-b^2)/(2b)`

`V=(1)/(2)(18b-b^3)`

Differentiate w. r.t b

`(dV)/(db)=(1)/(2)(18-3b^2)`

`(dV)/(db)=0`

`(1)/(2)(18-3b^2)=0`

`\implies 18-3b^2=0`

`\implies 3b^2=18`

`b^2=6`

`b=\pm √(6)`

`b=√(6)`

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

`(d^2V)/(db^2)=-3b`

At
`b=√(6)`

`(d^2V)/(db^2)=-3√(6)<0`

Hence, the volume of box is maximum at
`b=√(6)`.

`h=(18-(√(6))^2)/(2√(6))`

`h=(18-6)/(2√(6))`

`h=(12)/(2√(6))`

`h=√(6)`

`b=h=√(6)` m