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. 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

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