A) A rectangular pen is built with one side against a barn, 200 meters of fencing are used for the other three sides of the pen. What dimensions maximize the area of the pen?
B) A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of $100\space \text{m}^2$. What are the dimensions of each pen that minimize the amount of fence to be used?
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$\begingroup$A) Let the fence form a the $3$ sides of a rectangle; with the side of the barn being the $4^{th}$ side. Also let $x$ be the width of the rectangle and $y$ be its length. Clearly $2x+y=200 ....(1), \space \text{and the area enclosed (A) is given by:}\space A=xy....(2)$
Replacing $y$ in (2) by the expression in (1): $$\begin{align} A=xy=x(200-2x)= &200x-2x^2 \\ \frac{dA}{dx}=200-4x \\ \end{align}$$ To maximize the area, $\frac{dA}{dx}=0$ which is equivalent to $x=50 ; y=100$
$\endgroup$ 0 $\begingroup$B) Same situation here, except $A=4xy=400$ $$\begin{align} L=\text{total length of fence needed}&=5x+4y \\ \\ &= 5x+\frac{400}{x} \\ \frac{dL}{dx}=5-\frac{400}{x^2}\\ \end{align}$$ To minimize L, $\frac{dL}{dx}=0; \space \text{which is equivalent to} \space x=\sqrt{80}=4\sqrt{5} \approx 8.94; \text{and} \space y=5\sqrt{5} \approx 11.18 $
The question is not uniquely solvable because we don't know the barn's dimensions.
$\endgroup$ 4 $\begingroup$Derivatives are not needed for this problem.
The length restriction:$$2x+y=200$$The area restriction:$$A=x\cdot y$$Combining:$$2x+\frac{A}{x}=200$$ $$2x^2-200x+A=0$$The discriminant of this quadratic is:$$D=(-200)^2-8A$$Now, what is the maximum value of A that still results in a real solution?
The same procedure will solve Part B, with the appropriate area and length expressions, and considering the minimum $L$ that permits a real solution.
Edit to address comment;
For part B
The length restriction:$$5x+4y=L$$The area restriction:$$100=x\cdot y$$Combining (and multiplying through by $x$):$$5x+\frac{4\cdot 100}{x}=L$$ $$5x^2-Lx+400=0$$The discriminant of this quadratic is:$$D=(L)^2-4\cdot 5\cdot 400$$Now, what is the minimum value of L that still results in a real solution?
$\endgroup$ 2 $\begingroup$A) Using AM-GM inequality,
$$ \frac{(2x+y)^2}{4} \ge 2x \cdot y,\ \text{with equality in case} \ 2x = y $$
Plug in $2x +y = 200$, maximum area is $\frac{200^2}{8}$.
B) Similarly,$$ \frac{5x+4y}{2}\ge \sqrt{5x \cdot 4y},\ \text{with equality in case} \ 5x = 4y $$
Plug in $x\cdot y = 100$, minimal length is $\sqrt{5 \cdot 4 \cdot 100}$
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