# A fencing area question.

A farmer creates a rectangular pen by using one side of a barn as one side of the pen and using fencing for the other three sides. The farmer has $80~\text{ft}$ of fencing, and the side of the barn is $40~\text{ft}$ long. If $x$ represents the length of the fenced side of the pen that is parallel to the barn, then the length of each of the two fences sides of the pen that are perpendicular to the barn is $40 -.5x~\text{ft}$. For what values of $x$ is the area of the pen at least $600~\text{ft}$?

I know that they got $40-.5x$ by $x+y+y=80$. Thus, $x+2y=80$ and $y=40-5.x$ Area is then equal to $lw$, where $A=(40-.5x)(x)$?

Would it be times $40-2x$ since fence cannot be longer than barn at that side? or $40-x$?

How do I proceed to find area at least $600~\text{ft}$?

• What was the original question? – randomgirl Oct 16 '17 at 18:59
• $80-40=40$ so the shorter sides are 20 and area is 800. I don't understand what is $x$ and $y$ – Raffaele Oct 16 '17 at 19:20

So I guess you are asked to solve the inequality $$(40-.5x)x\geq 600$$ where the equality part will be recognized as the quadratic equation $$-.5x^2+40x-600=0$$ which has \begin{align} d&=b^2-4ac\\ &=40^2-4(-.5)(-600)\\ &=400 \end{align} so that the solutions are \begin{align} x&=\frac{-b\pm\sqrt d}{2a}\\ &=\frac{-40\pm 20}{2(-.5)}\\ &=40\pm 20 \end{align} and then when you realize that the graph of the quadratic expression is a parabola pointing downward, we see that the inequality is solved for $x\in[20,60]$.
NOTE: As N. F. Taussig correctly pointed out in the comments, the proper domain should be $0\leq x\leq 40$ and so the solution interval must be restricted to $[20,40]$.
• Your algebra is correct. However, if the side parallel to the barn is longer than the barn, the pen is open (not specifically prohibited by the question, but it does raise the question of the proper domain for $x$). – N. F. Taussig Oct 16 '17 at 19:47
You correctly found that $$A(x) = x\left(40 - \frac{1}{2}x\right)$$ Since we require that the area enclosed by the fence is at least $600~\text{ft}^2$ (you omitted the exponent),
\begin{align*} x\left(40 - \frac{1}{2}x\right) & \geq 600\\ 40x - \frac{1}{2}x^2 & \geq 600\\ x^2 - 80x & \leq -1200\\ x^2 - 80x + 1600 & \leq 400\\ (x - 40)^2 & \leq 400\\ |x - 40| & \leq 20 \end{align*} Thus, $x - 40 \leq 20$ and $x - 40 \geq -20$. \begin{align*} x - 40 & \leq 20 & x - 40 & \geq -20\\ x & \leq 60 & x & \geq 20 \end{align*} from which we conclude that $20 \leq x \leq 60$. That suggests that the side of the fence parallel to the barn should be between $20~\text{ft}$ and $60~\text{ft}$. However, if the side parallel to the barn exceeds $40~\text{ft}$, the pen cannot be closed, so the side parallel to the barn should be at least $20~\text{ft}$ and at most $40~\text{ft}$ long.