1. A manufacturer wants to design an open box having a square base and a surface area of $ \displaystyle 108$ square inches, as shown in figure . What dimensions will produce a box with maximum volume?
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$ \displaystyle \begin{array}{l}\text{ Total}\ \text{surface}\ \text{area =108 i}{{\text{n}}^{\text{2}}}\\\\\text{ (area of base)+(area of four sides) = 108}\\\\\ \ \ \ {{x}^{2}}+4xh=108\end{array}$
$ \displaystyle \ \ \ \ h=\frac{{108-{{x}^{2}}}}{{4x}}$
$ \displaystyle \begin{array}{l}\text{ Let the volume of the box be }V.\\\\\therefore \ \ V={{x}^{2}}h\end{array}$
$ \displaystyle \therefore \ \ V={{x}^{2}}\left( {\frac{{108-{{x}^{2}}}}{{4x}}} \right)=27x-\frac{1}{4}{{x}^{3}}$
$ \displaystyle \therefore \ \ \frac{{dV}}{{dx}}=27-\frac{3}{4}{{x}^{2}}$
$ \displaystyle \ \ \ \ \frac{{dV}}{{dx}}=0\ \text{when}$
$ \displaystyle \ \ \ \ 27-\frac{3}{4}{{x}^{2}}=0$
$ \displaystyle \therefore \ \ {{x}^{2}}=36\Rightarrow x=6\,\ \ \left[ {\because x>0} \right]$
$ \displaystyle \ \ \ \ \frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}=-\frac{3}{2}x$
$ \displaystyle \therefore \ \ {{\left. {\frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}} \right|}_{{x=6}}}=-\frac{3}{2}(6)=-9<0$
$ \displaystyle \therefore \ \ V\ \text{is maximum when }x=6\ \text{cm}\text{.}$
$ \displaystyle \therefore \ \ h=\frac{{108-{{6}^{2}}}}{{24}}=3\ \text{cm}$
$ \displaystyle \begin{array}{l}\therefore \ \ \text{The box has maximum volume when }\\\ \ \ \ x=6\ \text{cm and }h=3\ \text{cm}.\end{array}$
$ \displaystyle \ \ \ \ h=\frac{{108-{{x}^{2}}}}{{4x}}$
$ \displaystyle \begin{array}{l}\text{ Let the volume of the box be }V.\\\\\therefore \ \ V={{x}^{2}}h\end{array}$
$ \displaystyle \therefore \ \ V={{x}^{2}}\left( {\frac{{108-{{x}^{2}}}}{{4x}}} \right)=27x-\frac{1}{4}{{x}^{3}}$
$ \displaystyle \therefore \ \ \frac{{dV}}{{dx}}=27-\frac{3}{4}{{x}^{2}}$
$ \displaystyle \ \ \ \ \frac{{dV}}{{dx}}=0\ \text{when}$
$ \displaystyle \ \ \ \ 27-\frac{3}{4}{{x}^{2}}=0$
$ \displaystyle \therefore \ \ {{x}^{2}}=36\Rightarrow x=6\,\ \ \left[ {\because x>0} \right]$
$ \displaystyle \ \ \ \ \frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}=-\frac{3}{2}x$
$ \displaystyle \therefore \ \ {{\left. {\frac{{{{d}^{2}}V}}{{d{{x}^{2}}}}} \right|}_{{x=6}}}=-\frac{3}{2}(6)=-9<0$
$ \displaystyle \therefore \ \ V\ \text{is maximum when }x=6\ \text{cm}\text{.}$
$ \displaystyle \therefore \ \ h=\frac{{108-{{6}^{2}}}}{{24}}=3\ \text{cm}$
$ \displaystyle \begin{array}{l}\therefore \ \ \text{The box has maximum volume when }\\\ \ \ \ x=6\ \text{cm and }h=3\ \text{cm}.\end{array}$
2. Which points on the curve $ \displaystyle y = 4 − x^2$ are closest to the point $ \displaystyle (0, 2)?$
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$ \displaystyle \begin{array}{l}\text{ Let (}x,y)\ \text{be the required point on the curve}\text{.}\\\\\text{ Curve: }y=4-{{x}^{2}}\\\\\ \ \ \ \text{Let the distance between (0,2) and (}x,y)\ \text{be}\ s.\\\\\therefore \ \ \ s=\sqrt{{{{{(x-0)}}^{2}}+{{{(y-2)}}^{2}}}}\ \left[ {\text{using distance formula}} \right]\\\\\therefore \ \ \ s=\sqrt{{{{x}^{2}}+{{{(4-{{x}^{2}}-2)}}^{2}}}}\\\\\therefore \ \ \ s=\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}\end{array}$
$ \displaystyle \therefore \ \ \ \frac{{ds}}{{dx}}=\frac{{4{{x}^{3}}-6x}}{{2\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}$
$ \displaystyle \therefore \ \ \ \frac{{ds}}{{dx}}=\frac{{x(2{{x}^{2}}-3)}}{{\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}$
$ \displaystyle \ \ \ \ \ \frac{{ds}}{{dx}}=0\ \text{when}$
$ \displaystyle \ \ \ \ \ \frac{{x(2{{x}^{2}}-3)}}{{\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}=0$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Since }\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}\ne 0,\\\\\ \ \ \ \ x(2{{x}^{2}}-3)=0\end{array}$
$ \displaystyle \therefore \ \ \ x=0\ (\text{or) }x=-\sqrt{{\frac{3}{2}}}\ (\text{or)}\ x=\sqrt{{\frac{3}{2}}}\ $
$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}=\frac{{2{{x}^{6}}-9{{x}^{4}}+24{{x}^{2}}-12}}{{{{{\left( {{{x}^{4}}-3{{x}^{2}}+4} \right)}}^{{\frac{3}{2}}}}}}$
$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=0}}}=-\frac{3}{2}<0$
$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=-\sqrt{{\frac{3}{2}}}}}}=\frac{{12}}{{\sqrt{7}}}>0$
$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=\sqrt{{\frac{3}{2}}}}}}=\frac{{12}}{{\sqrt{7}}}>0$
$ \displaystyle \therefore \ \ s\text{ is maximum volume when }$
$ \displaystyle \ \ \ \ x=-\sqrt{{\frac{3}{2}}}\text{ and }x=-\sqrt{{\frac{3}{2}}}.$
$ \displaystyle \therefore \ \ \text{When }x=-\sqrt{{\frac{3}{2}}},y=\frac{5}{2}.$
$ \displaystyle \ \ \ \ \text{When }x=\sqrt{{\frac{3}{2}}},y=\frac{5}{2}$
$ \displaystyle \therefore \ \ \text{The closest ponits on the curve are}$
$ \displaystyle \ \ \ \ \left( {-\sqrt{{\frac{3}{2}}},\frac{5}{2}} \right)\text{and }\left( {\sqrt{{\frac{3}{2}}},\frac{5}{2}} \right).$
$ \displaystyle \therefore \ \ \ \frac{{ds}}{{dx}}=\frac{{4{{x}^{3}}-6x}}{{2\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}$
$ \displaystyle \therefore \ \ \ \frac{{ds}}{{dx}}=\frac{{x(2{{x}^{2}}-3)}}{{\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}$
$ \displaystyle \ \ \ \ \ \frac{{ds}}{{dx}}=0\ \text{when}$
$ \displaystyle \ \ \ \ \ \frac{{x(2{{x}^{2}}-3)}}{{\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}}}=0$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Since }\sqrt{{{{x}^{4}}-3{{x}^{2}}+4}}\ne 0,\\\\\ \ \ \ \ x(2{{x}^{2}}-3)=0\end{array}$
$ \displaystyle \therefore \ \ \ x=0\ (\text{or) }x=-\sqrt{{\frac{3}{2}}}\ (\text{or)}\ x=\sqrt{{\frac{3}{2}}}\ $
$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}=\frac{{2{{x}^{6}}-9{{x}^{4}}+24{{x}^{2}}-12}}{{{{{\left( {{{x}^{4}}-3{{x}^{2}}+4} \right)}}^{{\frac{3}{2}}}}}}$
$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=0}}}=-\frac{3}{2}<0$
$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=-\sqrt{{\frac{3}{2}}}}}}=\frac{{12}}{{\sqrt{7}}}>0$
$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}s}}{{d{{x}^{2}}}}} \right|}_{{x=\sqrt{{\frac{3}{2}}}}}}=\frac{{12}}{{\sqrt{7}}}>0$
$ \displaystyle \therefore \ \ s\text{ is maximum volume when }$
$ \displaystyle \ \ \ \ x=-\sqrt{{\frac{3}{2}}}\text{ and }x=-\sqrt{{\frac{3}{2}}}.$
$ \displaystyle \therefore \ \ \text{When }x=-\sqrt{{\frac{3}{2}}},y=\frac{5}{2}.$
$ \displaystyle \ \ \ \ \text{When }x=\sqrt{{\frac{3}{2}}},y=\frac{5}{2}$
$ \displaystyle \therefore \ \ \text{The closest ponits on the curve are}$
$ \displaystyle \ \ \ \ \left( {-\sqrt{{\frac{3}{2}}},\frac{5}{2}} \right)\text{and }\left( {\sqrt{{\frac{3}{2}}},\frac{5}{2}} \right).$
3. A rectangular page is to contain $ \displaystyle 24$ square inches of print. The margins at the top and bottom of the page are to be $ \displaystyle 1 \frac{1}{2}$ inches, and the margins on the left and right are to be $ \displaystyle 1$ inch (see Figure). What should the dimensions of the page be so that the least amount of paper is used?
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$ \displaystyle \begin{array}{l} \ \ \ \text{Let the length and width of the printed }\\\ \ \ \text{area be }x\ \text{and }y\text{ respectively}\text{.}\\\\\text{ By the problem, }xy=24\end{array}$
$ \displaystyle \therefore \ \ y=\frac{{24}}{x}$
$ \displaystyle \ \ \ \ \text{Let the area of the paper needed be }A.$
$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {A=(x+3)(y+2)} \end{array}$
$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {A=(x+3)(\frac{{24}}{x}+2)} \end{array}=30+2x+\frac{{72}}{x}$
$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2-\frac{{72}}{{{{x}^{2}}}}$
$ \displaystyle \ \ \ \ \ \frac{{dA}}{{dx}}=0\ \text{when }2-\frac{{72}}{{{{x}^{2}}}}=0.$
$ \displaystyle \begin{array}{l}\therefore \ \ \ {{x}^{2}}=36\\\\\ \ \ \ \ x=6\ \ \ (\because x>0)\end{array}$
$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}=\frac{{144}}{{{{x}^{3}}}}$
$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}} \right|}_{{x=6}}}=\frac{{144}}{{{{6}^{3}}}}>0$
$ \displaystyle \therefore \ \ A\text{ is maximum volume when }x=6$
$ \displaystyle \therefore \ \ \text{ }y=\frac{{24}}{6}=4\ \text{cm}.$
$ \displaystyle \therefore \ \ \text{The dimenssions of the paper}=4\ \text{cm}\times 6\text{cm}.$
$ \displaystyle \therefore \ \ y=\frac{{24}}{x}$
$ \displaystyle \ \ \ \ \text{Let the area of the paper needed be }A.$
$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {A=(x+3)(y+2)} \end{array}$
$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {A=(x+3)(\frac{{24}}{x}+2)} \end{array}=30+2x+\frac{{72}}{x}$
$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2-\frac{{72}}{{{{x}^{2}}}}$
$ \displaystyle \ \ \ \ \ \frac{{dA}}{{dx}}=0\ \text{when }2-\frac{{72}}{{{{x}^{2}}}}=0.$
$ \displaystyle \begin{array}{l}\therefore \ \ \ {{x}^{2}}=36\\\\\ \ \ \ \ x=6\ \ \ (\because x>0)\end{array}$
$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}=\frac{{144}}{{{{x}^{3}}}}$
$ \displaystyle \ \ \ \ \ {{\left. {\frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}} \right|}_{{x=6}}}=\frac{{144}}{{{{6}^{3}}}}>0$
$ \displaystyle \therefore \ \ A\text{ is maximum volume when }x=6$
$ \displaystyle \therefore \ \ \text{ }y=\frac{{24}}{6}=4\ \text{cm}.$
$ \displaystyle \therefore \ \ \text{The dimenssions of the paper}=4\ \text{cm}\times 6\text{cm}.$
4. Two posts, one $ \displaystyle 12$ feet high and the other $ \displaystyle 28$ feet high, stand $ \displaystyle 30$ feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. Where should the stake be placed to use the least amount of wire?
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$ \displaystyle \begin{array}{l} \ \ \ \text{Let the length of wire required to attach 12-foot post be }y\ \ \ \\\ \ \ \ \text{and that required to attach 28-foot post be }z\text{.}\\\\\ \ \ \text{Let the distance of the stack from 12-foot post be }x.\\\\\therefore \ \ {{x}^{2}}+{{12}^{2}}={{y}^{2}}\Rightarrow y=\sqrt{{144+{{x}^{2}}}}\\\\\ \ \ \ {{(30-x)}^{2}}+{{28}^{2}}={{z}^{2}}\Rightarrow z=\sqrt{{{{x}^{2}}-60x+1684}}.\\\\\ \ \ \ \text{Let the total length of wire required be }W.\end{array}$
$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {W=x+y} \end{array}=\sqrt{{144+{{x}^{2}}}}+\sqrt{{{{x}^{2}}-60x+1684}}.$
$ \displaystyle \therefore \ \ \ \frac{{dW}}{{dx}}=\frac{x}{{\sqrt{{144+{{x}^{2}}}}}}+\frac{{x-30}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}$
$ \displaystyle \ \ \ \ \ \frac{{dW}}{{dx}}=0\ \text{when }\frac{x}{{\sqrt{{144+{{x}^{2}}}}}}+\frac{{x-30}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}=0.$
$ \displaystyle \therefore \ \ \ \frac{x}{{\sqrt{{144+{{x}^{2}}}}}}=\frac{{30-x}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}$
$ \displaystyle \begin{array}{l}\ \ \ \ \ x\sqrt{{{{x}^{2}}-60x+1684}}=(30-x)\sqrt{{{{x}^{2}}+144}}\\\\\ \ \ \ \ {{x}^{2}}({{x}^{2}}-60x+1684)={{(30-x)}^{2}}(x2+144)\\\\\ \ \ \ \ {{x}^{4}}-60{{x}^{3}}+1684{{x}^{2}}={{x}^{4}}-60{{x}^{3}}+1044{{x}^{2}}-8640x+129,600\\\\\ \ \ \ \ 640{{x}^{2}}+8640x-129,600=0\\\\\ \ \ \ \ 2{{x}^{2}}+27x-405=0\\\\\ \ \ \ \ (x-9)(2x+45)=0\\\\\ \ \ \ \ \text{Since }x>0,\ x=9\\\\\ \ \ \ \ W=\sqrt{{144+{{x}^{2}}}}+\sqrt{{{{x}^{2}}-60x+1684}}.\\\\\therefore \ \ \ \text{When }x=0,\\\\\ \ \ \ \ W=\sqrt{{144}}+\sqrt{{1684}}\\\\\ \ \ \ \ \ \ \ \ \ \ =12+41.04=53.04\\\\\ \ \ \ \ \text{When }x=9,\\\\\ \ \ \ \ W=\sqrt{{144+81}}+\sqrt{{81-540+1684}}\\\\\ \ \ \ \ \ \ \ \ =15+35=50\\\\\ \ \ \ \ \text{When }x=30,\\\\\ \ \ \ \ W=\sqrt{{144+900}}+\sqrt{{900-1800+1684}}\\\\\ \ \ \ \ \ \ \ \ \ \ =32.31+28=60.31\\\\\therefore \ \ \ W\text{ is maximum when }x=9.\\\\\therefore \ \ \ \text{The}\ \text{wires}\ \text{should}\ \text{be}\ \text{staked}\ \text{at}\ \text{9}\ \text{feet}\ \text{from}\ \text{the}\ \text{12-foot}\ \text{pole}.\end{array}$
$ \displaystyle \therefore \ \ \ \begin{array}{*{20}{l}} {W=x+y} \end{array}=\sqrt{{144+{{x}^{2}}}}+\sqrt{{{{x}^{2}}-60x+1684}}.$
$ \displaystyle \therefore \ \ \ \frac{{dW}}{{dx}}=\frac{x}{{\sqrt{{144+{{x}^{2}}}}}}+\frac{{x-30}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}$
$ \displaystyle \ \ \ \ \ \frac{{dW}}{{dx}}=0\ \text{when }\frac{x}{{\sqrt{{144+{{x}^{2}}}}}}+\frac{{x-30}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}=0.$
$ \displaystyle \therefore \ \ \ \frac{x}{{\sqrt{{144+{{x}^{2}}}}}}=\frac{{30-x}}{{\sqrt{{{{x}^{2}}-60x+1684}}}}$
$ \displaystyle \begin{array}{l}\ \ \ \ \ x\sqrt{{{{x}^{2}}-60x+1684}}=(30-x)\sqrt{{{{x}^{2}}+144}}\\\\\ \ \ \ \ {{x}^{2}}({{x}^{2}}-60x+1684)={{(30-x)}^{2}}(x2+144)\\\\\ \ \ \ \ {{x}^{4}}-60{{x}^{3}}+1684{{x}^{2}}={{x}^{4}}-60{{x}^{3}}+1044{{x}^{2}}-8640x+129,600\\\\\ \ \ \ \ 640{{x}^{2}}+8640x-129,600=0\\\\\ \ \ \ \ 2{{x}^{2}}+27x-405=0\\\\\ \ \ \ \ (x-9)(2x+45)=0\\\\\ \ \ \ \ \text{Since }x>0,\ x=9\\\\\ \ \ \ \ W=\sqrt{{144+{{x}^{2}}}}+\sqrt{{{{x}^{2}}-60x+1684}}.\\\\\therefore \ \ \ \text{When }x=0,\\\\\ \ \ \ \ W=\sqrt{{144}}+\sqrt{{1684}}\\\\\ \ \ \ \ \ \ \ \ \ \ =12+41.04=53.04\\\\\ \ \ \ \ \text{When }x=9,\\\\\ \ \ \ \ W=\sqrt{{144+81}}+\sqrt{{81-540+1684}}\\\\\ \ \ \ \ \ \ \ \ =15+35=50\\\\\ \ \ \ \ \text{When }x=30,\\\\\ \ \ \ \ W=\sqrt{{144+900}}+\sqrt{{900-1800+1684}}\\\\\ \ \ \ \ \ \ \ \ \ \ =32.31+28=60.31\\\\\therefore \ \ \ W\text{ is maximum when }x=9.\\\\\therefore \ \ \ \text{The}\ \text{wires}\ \text{should}\ \text{be}\ \text{staked}\ \text{at}\ \text{9}\ \text{feet}\ \text{from}\ \text{the}\ \text{12-foot}\ \text{pole}.\end{array}$
5. The sum of the perimeters of a circle and square is $ \displaystyle k,$ where $ \displaystyle k$ is some constant. Prove that the sum of their areas is least, when the side of the square is double the radius of the circle.
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$ \displaystyle \therefore \ \ \ r=~\frac{{k-4x}}{{2\pi }}$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Let the total are be }A.\\\\\therefore \ \ \ A={{x}^{2}}+\pi {{r}^{2}}\end{array}$
$ \displaystyle \therefore \ \ \ A={{x}^{2}}+\pi {{\left( {\frac{{k-4x}}{{2\pi }}} \right)}^{2}}$
$ \displaystyle \therefore \ \ \ A={{x}^{2}}+\frac{{{{{(k-4x)}}^{2}}}}{{4\pi }}$
$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2x+\frac{{2(-4)(k-4x)}}{{4\pi }}$
$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=2\left( {x+\frac{{4x-k}}{\pi }} \right)$
$ \displaystyle \therefore \ \ \ \frac{{dA}}{{dx}}=\frac{2}{\pi }\left[ {(\pi +4)x-k} \right]$
$ \displaystyle \ \ \ \ \ \frac{{dA}}{{dx}}=0\ \text{when}\ \frac{2}{\pi }\left[ {(\pi +4)x-k} \right]=0$
$ \displaystyle \therefore \ \ \ x=\frac{k}{{\pi +4}}$
$ \displaystyle \ \ \ \ \ \frac{{{{d}^{2}}A}}{{d{{x}^{2}}}}=\frac{{2(\pi +4)}}{\pi }>0$
$ \displaystyle \therefore \ \ \ A\ \text{is minimum when }x=\frac{k}{{\pi +4}}.$
$ \displaystyle \therefore \ \ \ r=\frac{1}{{2\pi }}[k-\frac{{4k}}{{\pi +4}}]$
$ \displaystyle \ \ \ \ \ \ \ \ =\frac{{\pi k+4k-4k}}{{2\pi (\pi +4)}}$
$ \displaystyle \ \ \ \ \ \ \ \ =~\frac{1}{2}\left( {\frac{k}{{\pi +4}}} \right)$
$ \displaystyle \ \ \ \ \ \ \ \ =~\frac{1}{2}x$
$ \displaystyle \begin{array}{l}\therefore \ \ x=2r\\\\\ \ \ \text{Hence the sum of their areas is least},\\\ \ \ \text{when the side of the square is double }\\\ \ \ \text{the radius of the circle}.\end{array}$
6. Find two positive numbers that the sum of the first number cubed and the second number is $ \displaystyle 500$ and the product is a maximum.
7. An open box of maximum volume is to be made from a square piece of material, $ \displaystyle 24$ inches on a side, by cutting equal squares from the corners and turning up the sides (see figure). How much material should be cut?
8. A Norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window (see figure). Find the dimensions of a Norman window of maximum area when the total perimeter is $ \displaystyle 16$ feet.
9. Find the area of the largest isosceles triangle that can be inscribed in a circle of radius $ \displaystyle 6$ (see figure).
(a) Solve by writing the area as a function of $ \displaystyle h$.
(b) Solve by writing the area as a function of $ \displaystyle α$.
(c) Identify the type of triangle of maximum area.
(a) Solve by writing the area as a function of $ \displaystyle h$.
(b) Solve by writing the area as a function of $ \displaystyle α$.
(c) Identify the type of triangle of maximum area.
10. Twenty feet of wire is to be used to form two figures: equilateral triangle and square, how much wire should be used for each figure so that the total enclosed area is maximum?
11. A right triangle is formed in the first quadrant by the $ \displaystyle x$- and $ \displaystyle y$-axes and a line through the point $ \displaystyle (1, 2)$ (see figure).
(a) Write the length $ \displaystyle L$ of the hypotenuse as a function of $ \displaystyle x.$
(b) Find the vertices of the triangle such that its area is a minimum.
(a) Write the length $ \displaystyle L$ of the hypotenuse as a function of $ \displaystyle x.$
(b) Find the vertices of the triangle such that its area is a minimum.
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