# 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 leng

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

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1. thnahdat

$$b=h=\sqrt{6}$$ m

Step-by-step 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=$$2b^2+4bh$$

$$2b^2+4bh=36$$

$$b^2+2bh=18$$

$$2bh=18-b^2$$

$$h=\frac{18-b^2}{2b}$$

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

Substitute the values

$$V=b^2\times \frac{18-b^2}{2b}$$

$$V=\frac{1}{2}(18b-b^3)$$

Differentiate w. r.t b

$$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$$

$$\frac{dV}{db}=0$$

$$\frac{1}{2}(18-3b^2)=0$$

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

$$\implies 3b^2=18$$

$$b^2=6$$

$$b=\pm \sqrt{6}$$

$$b=\sqrt{6}$$

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

Again differentiate w.r.t b

$$\frac{d^2V}{db^2}=-3b$$

At  $$b=\sqrt{6}$$

$$\frac{d^2V}{db^2}=-3\sqrt{6}<0$$

Hence, the volume of box is maximum at $$b=\sqrt{6}$$.

$$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$$

$$h=\frac{18-6}{2\sqrt{6}}$$

$$h=\frac{12}{2\sqrt{6}}$$

$$h=\sqrt{6}$$

$$b=h=\sqrt{6}$$ m