91Ó°ÊÓ

The bottom will set you back $\$2$ per square foot, the sides will set you back $\$5$ per square foot, and the top will set you back $\$7$ per square foot. The box has a volume of $20$ cubic feet.

Let $x$be the length, $y$be the width and $z$be the height of the rectangular box.

The goal is to determine the rectangle's dimension that will reduce the box's cost.

The volume $V$of the box is given by

$$\begin{gathered} V=xyz=20 \\ z=\frac{20}{xy}…….\left(1\right) \end{gathered}$$

The plywood used for the box's bottom cost $\$2xy$

The pine for the box's four sides was $2(\$5)xz+2(\$5)yz$or $\$10xz+\$10yz$

The oak for the top of the box cost $\$7xy$

Thus, the total cost is

$$\begin{gathered} C(x,y)=2xy+10xz+10yz+7xy \\ =9xy+10x·\frac{20}{xy}+10y·\frac{20}{xy}\left[\text{Since}z=\frac{20}{xy}\right.\text{]} \\ =9xy+\frac{200}{y}+\frac{200}{x}…….\left(2\right) \end{gathered}$$

Differentiate $\left(2\right)$partially with respect to $x$

$$\begin{gathered} \frac{∂}{∂x}C(x,y)=\frac{∂}{∂x}\left(9xy+\frac{200}{y}+\frac{200}{x}\right) \\ {C}_x(x,y)=9y\frac{∂}{∂x}x+\frac{∂}{∂x}\frac{200}{y}+\frac{∂}{∂x}\frac{200}{x} \\ =9y+0-\frac{200}{x^2} \\ =9y-\frac{200}{x^2}⋯.. \end{gathered}$$

Differentiate ( 2 ) partially with respect to $y$

$$\begin{gathered} \frac{∂}{∂y}C(x,y)=\frac{∂}{∂y}\left\{9xy+\frac{200}{y}+\frac{200}{x}\right\} \\ {C}_y(x,y)=9x\frac{∂}{∂y}y+\frac{∂}{∂y}\frac{200}{y}+\frac{∂}{∂y}\frac{200}{x} \\ =9x-\frac{200}{y^2}+0 \\ =9x-\frac{200}{y^2}….\left(4\right) \end{gathered}$$

The stationary point is given by

$C_{,}(x,y)=0\text{and}{C}_s(x,y)=0$

Thus,

$$\begin{gathered} C_x(x,y)=0 \\ 9y-\frac{200}{x^2}=0 \end{gathered}$$

And

$$\begin{gathered} C,(x,y)=0 \\ 9x-\frac{200}{y^2}=0……\left(6\right) \end{gathered}$$

$step4$of $70\backslash \mathrm{AA}$

Now, solve for $x$and $y$

Multiply$\left(5\right)$by $x$and multiply $\left(6\right)$by $y$and then subtract.

$x\left(9y-\frac{200}{x^2}\right)-y\left(9x-\frac{200}{y^2}\right)=0$

$9xy-\frac{200}{x}-9xy+\frac{200}{y}=0$

$-\frac{200}{x}+\frac{200}{y}=0$

$\frac{200}{x}=\frac{200}{y}$

$\frac{200}{x}=\frac{200}{y}$

$x=y$

$9xy-\frac{200}{x}-9xy+\frac{200}{y}=0$

Put $y=x$in$(5)$

$$\begin{gathered} 9x-\frac{200}{x^2}=0 \\ 9x=\frac{200}{x^2} \\ 9x^3=200 \\ {x}^3=\frac{200}{9} \end{gathered}$$

$$\begin{gathered} x=\sqrt[3]{\frac{200}{9}} \\ =\sqrt[3]{\frac{200·5·81}{9·81·5}} \\ =\sqrt[3]{\frac{1000·81}{9^3·5}} \\ =\frac{10}{9}\sqrt[3]{\frac{81}{5}} \end{gathered}$$

Since $x=y$so $y=\frac{10}{9}\sqrt[3]{\frac{81}{5}}$

To determine the nature point calculate $C_{zx}(x,y),C_{yy}(x,y)$and $C_{xy}(x,y)$Differentiate ( 3 ) partially with respect to $x$

$$\begin{gathered} \frac{∂}{∂x}{C}_x(x,y)=\frac{∂}{∂x}\left(9y-\frac{200}{x^2}\right) \\ {C}_{xx}(x,y)=9\frac{∂}{∂x}y-\frac{∂}{∂x}\frac{200}{x^2} \\ =0+2·200·\frac{1}{x^3} \\ =\frac{400}{x^3} \end{gathered}$$

Partially differentiate $\left(4\right)$with regard to $y$

$$\begin{gathered} \frac{∂}{∂y}{C}_y(x,y)=\frac{∂}{∂y}\left(9x-\frac{200}{y^2}\right) \\ {C}_{yy}(x,y)=9\frac{∂}{∂y}x-\frac{∂}{∂y}\frac{200}{y^2} \\ =0+2·200·\frac{1}{y^3} \\ =\frac{400}{y^3} \end{gathered}$$

Differentiate $(3)$partially with respect to $y$

$$\begin{gathered} \frac{∂}{∂y}{C}_x(x,y)=\frac{∂}{∂y}\left(9y-\frac{200}{x^2}\right) \\ {C}_{xy}(x,y)=9\frac{∂}{∂y}y-\frac{∂}{∂y}\frac{200}{x^2} \end{gathered}$$

$$\begin{gathered} =9-0 \\ =9 \end{gathered}$$Now, find the discriminant of $C(x,y)$at $\left(\sqrt[3]{\frac{200}{9}},\sqrt[3]{\frac{200}{9}}\right)$$$\begin{gathered} det\left(H_C\right)=C_{xx}{C}_{yy}-{\left(C_{xy}\right)}^2 \\ =\frac{400}{x^3}·\frac{400}{y^3}-9^2 \\ =\frac{400}{{\left(\sqrt[3]{\frac{200}{9}}\right)}^3}·\frac{400}{{\left(\sqrt[3]{\frac{200}{9}}\right)}^3}-81 \\ =\frac{400}{\left(\frac{200}{9}\right)}·\frac{400}{\left(\frac{200}{9}\right)}-81 \\ =400·\frac{9}{200}·400·\frac{9}{200}-81 \\ =2·9·2·9-81 \\ =324-81 \\ =243 \end{gathered}$$

$$\begin{gathered} \left(\sqrt[3]{\frac{200}{9}},\sqrt[3]{\frac{200}{9}}\right) \\ {C}_x(x,y)=\frac{400}{{\left(\sqrt[3]{\frac{200}{9}}\right)}^3} \\ =\frac{400}{\left(\frac{200}{9}\right)} \\ =400·\frac{9}{200} \\ =18 \end{gathered}$$

The function $C(x,y)$will have a relative minimum if $det\left(H_C\right)>0$with $C_x(x,y)>0$or $C_{yy}(x,y)>0$

Since $det\left(H_C\right)=243>0$and $C_{xx}(x,y)=18>0$so the function has a minimum at $\left(\sqrt[3]{\frac{200}{9}},\sqrt[3]{\frac{200}{9}}\right)$i.e. $\left(\frac{10}{9}\sqrt[3]{\frac{81}{5}},\frac{10}{9}\sqrt[3]{\frac{81}{5}}\right)$

Put $x=y=\sqrt{\frac{200}{9}}$ in $\left(1\right)$ to find $z$

$z=\frac{20}{\sqrt{\frac{200}{9}}·\sqrt{\frac{200}{9}}}$

$=\frac{20}{\sqrt{\frac{200}{9}+\frac{200}{9}}}$

$=\frac{20}{\sqrt{\frac{40}{3}\frac{1000}{27}}}$

$=\frac{20}{\frac{10}{3}\sqrt[3]{\frac{40}{3}}}$

$=\frac{60}{10\sqrt{\frac{40}{3}}}$

$=\frac{6}{\sqrt{\frac{40}{3}}}$

$=\frac{6\sqrt[3]{\frac{3}{40}}}{\sqrt{\frac{40}{3}}·\sqrt[3]{\frac{3}{40}}}$

$=6·\sqrt{\frac{3}{40}}$

$=\sqrt[3]{\frac{3·216}{40}}$

$=\sqrt[3]{\frac{3-27}{5}}$

$=\sqrt[3]{\frac{81}{5}}$

Therefore, the side of the rectangle are $\frac{10}{9}\sqrt[3]{\frac{81}{5}}$ feet $\frac{10}{9}\sqrt{\frac{81}{5}}$feet and the height of the rectangle is $\sqrt[3]{\frac{81}{5}}$ feet.