Answer:
[tex]b=h=\sqrt{6}[/tex] 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=[tex]2b^2+4bh[/tex]
[tex]2b^2+4bh=36[/tex]
[tex]b^2+2bh=18[/tex]
[tex]2bh=18-b^2[/tex]
[tex]h=\frac{18-b^2}{2b}[/tex]
Volume of box, V=[tex]b^2h[/tex]
Substitute the values
[tex]V=b^2\times \frac{18-b^2}{2b}[/tex]
[tex]V=\frac{1}{2}(18b-b^3)[/tex]
Differentiate w. r.t b
[tex]\frac{dV}{db}=\frac{1}{2}(18-3b^2)[/tex]
[tex]\frac{dV}{db}=0[/tex]
[tex]\frac{1}{2}(18-3b^2)=0[/tex]
[tex]\implies 18-3b^2=0[/tex]
[tex]\implies 3b^2=18[/tex]
[tex]b^2=6[/tex]
[tex]b=\pm \sqrt{6}[/tex]
[tex]b=\sqrt{6}[/tex]
The negative value of b is not possible because length cannot be negative.
Again differentiate w.r.t b
[tex]\frac{d^2V}{db^2}=-3b[/tex]
At [tex]b=\sqrt{6}[/tex]
[tex]\frac{d^2V}{db^2}=-3\sqrt{6}<0[/tex]
Hence, the volume of box is maximum at [tex]b=\sqrt{6}[/tex].
[tex]h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}[/tex]
[tex]h=\frac{18-6}{2\sqrt{6}}[/tex]
[tex]h=\frac{12}{2\sqrt{6}}[/tex]
[tex]h=\sqrt{6}[/tex]
[tex]b=h=\sqrt{6}[/tex] m
