Respuesta :
Answer:
The dimensions of the pen that minimize the cost of fencing are:
[tex]x \approx 12.25 \:ft[/tex]
[tex]y \approx 8.17 \:ft[/tex]
Step-by-step explanation:
Let [tex]x[/tex] be the width and [tex]y[/tex] the length of the rectangular pen.
We know that the area of this rectangle is going to be [tex]x\cdot y[/tex].The problem tells us that the area is 100 feet, so we get the constraint equation:
[tex]x\cdot y=100[/tex]
The quantity we want to optimize is going to be the cost to make our fence. If we have chain-link on three sides of the pen, say one side of length [tex]y[/tex] and both sides of length [tex]x[/tex], the cost for these sides will be
[tex]10(y+2x)[/tex]
and the remaining side will be fence and hence have cost
[tex]20y[/tex]
Thus we have the objective equation:
[tex]C=10(y+2x)+20y\\C=10y+20x+20y\\C=30y+20x[/tex]
We can solve the constraint equation for one of the variables to get:
[tex]x\cdot y=100\\y=\frac{100}{x}[/tex]
Thus, we get the cost equation in terms of one variable:
[tex]C=30(\frac{100}{x})+20x\\C=\frac{3000}{x}+20x[/tex]
We want to find the dimensions that minimize the cost of the pen, for this reason, we take the derivative of the cost equation and set it equal to zero.
[tex]\frac{d}{dx} C=\frac{d}{dx} (\frac{3000}{x}+20x)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\C'(x)=\frac{d}{dx}\left(\frac{3000}{x}\right)+\frac{d}{dx}\left(20x\right)\\\\C'(x)=-\frac{3000}{x^2}+20[/tex]
[tex]C'(x)=-\frac{3000}{x^2}+20=0\\\\-\frac{3000}{x^2}x^2+20x^2=0\cdot \:x^2\\-3000+20x^2=0\\-3000+20x^2+3000=0+3000\\20x^2=3000\\\frac{20x^2}{20}=\frac{3000}{20}\\x^2=150\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{150},\:x=-\sqrt{150}[/tex]
Because length must always be zero or positive we take [tex]x=\sqrt{150}[/tex] as only value for the width.
To check that this is indeed a value of [tex]x[/tex] that gives us a minimum, we need to take the second derivative of our cost function.
[tex]\frac{d}{dx} C'(x)=\frac{d}{dx} (-\frac{3000}{x^2}+20)\\\\C''(x)=-\frac{d}{dx}\left(\frac{3000}{x^2}\right)+\frac{d}{dx}\left(20\right)\\\\C''(x)=\frac{6000}{x^3}[/tex]
Because [tex]C''(\sqrt{150})=\frac{6000}{\left(\sqrt{150}\right)^3}=\frac{4\sqrt{6}}{3}[/tex] is greater than zero, [tex]x=\sqrt{150}[/tex] is a minimum.
Now, we need values of both x and y, thus as [tex]y=\frac{100}{x}[/tex], we get
[tex]x=\sqrt{150}=5\sqrt{6}=12.25[/tex]
[tex]y=\frac{100}{\sqrt{150}}=\frac{10\sqrt{6}}{3}\approx 8.17[/tex]
The dimensions of the pen that minimize the cost of fencing are:
[tex]x \approx 12.25 \:ft[/tex]
[tex]y \approx 8.17 \:ft[/tex]
