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katen-ka-za [31]
3 years ago
9

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

nd base length b of the box be so as to maximize its volume
Mathematics
1 answer:
kkurt [141]3 years ago
7 0

Answer:

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}

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

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