Question

We want to construct a box with a square base and we currently only have 10m2 of material to use in construction of the box. Assuming that all material is used in the construction process, determine the maximum volume that the box can have.

Answer 1

Answer:

The maximum volume of the box is:

$V =\frac{5}{3}\sqrt{\frac{5}{3}}$

Step-by-step explanation:

Given

$Surface\ Area = 10m^2$

Required

The maximum volume of the box

Let

$a \to base\ dimension$

$b \to height$

The surface area of the box is:

$Surface\ Area = 2(a*a + a*b + a*b)$

$Surface\ Area = 2(a^2 + ab + ab)$

$Surface\ Area = 2(a^2 + 2ab)$

So, we have:

$2(a^2 + 2ab) = 10$

Divide both sides by 2

$a^2 + 2ab = 5$

Make b the subject

$2ab = 5 -a^2$

$b = \frac{5 -a^2}{2a}$

The volume of the box is:

$V = a*a*b$

$V = a^2b$

Substitute: $b = \frac{5 -a^2}{2a}$

$V = a^2*\frac{5 - a^2}{2a}$

$V = a*\frac{5 - a^2}{2}$

$V = \frac{5a - a^3}{2}$

Spit

$V = \frac{5a}{2} - \frac{a^3}{2}$

Differentiate V with respect to a

$V' = \frac{5}{2} -3 * \frac{a^2}{2}$

$V' = \frac{5}{2} -\frac{3a^2}{2}$

Set $V' =0$ to calculate a

$0 = \frac{5}{2} -\frac{3a^2}{2}$

Collect like terms

$\frac{3a^2}{2} = \frac{5}{2}$

Multiply both sides by 2

$3a^2= 5$

Solve for a

$a^2= \frac{5}{3}$

$a= \sqrt{\frac{5}{3}}$

Recall that:

$b = \frac{5 -a^2}{2a}$

$b = \frac{5 -(\sqrt{\frac{5}{3}})^2}{2*\sqrt{\frac{5}{3}}}$

$b = \frac{5 -\frac{5}{3}}{2*\sqrt{\frac{5}{3}}}$

$b = \frac{\frac{15 - 5}{3}}{2*\sqrt{\frac{5}{3}}}$

$b = \frac{\frac{10}{3}}{2*\sqrt{\frac{5}{3}}}$

$b = \frac{\frac{5}{3}}{\sqrt{\frac{5}{3}}}$

Apply law of indices

$b = (\frac{5}{3})^{1 - \frac{1}{2}}$

$b = (\frac{5}{3})^{\frac{1}{2}}$

$b = \sqrt{\frac{5}{3}}$

So:

$V = a^2b$

$V =\sqrt{(\frac{5}{3})^2} * \sqrt{\frac{5}{3}}$

$V =\frac{5}{3} * \sqrt{\frac{5}{3}}$

$V =\frac{5}{3}\sqrt{\frac{5}{3}}$