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Ella Mental has $600$ ft of fencing to enclose two fields. One is to be a rectangle twice as long as it is wide and the other is to be a square. The square field must contain at least $100$ ft squared. The rectangular one must contain at least $800$ ft squared.

a. If $x$ is the width of the rectangular field, what is the domain of $x$?

b. Plot the graph of the total area contained in the two fields as a function of $x$.

c. What is the greatest area that can be contained in the two fields? Justify your answer

By the way, the answers to a, b, c are...(according to the textbook)

a. domain: $20\le x\le 93.333333\dots$

b. $A(x) = 22500 -450x + 4.25x$ squared

c. greatest area $= 17522.2222$

I keep getting the wrong answer for the greatest area. Please provide me with explanations to each

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    $\begingroup$ The maximum is obtained at one of the endpoints of the domain. (You need to check these as well as values at any critical points.) $\endgroup$ – David Mitra Aug 9 '12 at 20:39
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In general, for a maximum, the candidates are (i) where the derivative is $0$ (or doesn't exist) and (ii) the endpoints. Check endpoints. And for this problem, don't bother calculating the derivative! It is obvious that the maximum must be at an endpoint, since we have an upward facing parabola.

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Since $x$ is the length (in feet) of the short side of the rectangular field, then the area of the rectangular field is $2x^2$, and its perimeter is $6x$. Thus, the side length of the square field is $(600-6x)/4=150-\frac32x$, so its area is $22500-450x+\frac94x^2$, and the total area is $$A(x)=22500-450x+\frac{17}4x^2,$$ as desired.

We know in particular that we need $2x^2\geq 800$, from which $x\geq 20$ (since $x$ is positive). Also, since the square field needs area at least $100$ square feet, then its side length is at least $10$ feet, meaning $150-\frac32x\geq 10,$ from which $x\leq\frac{280}3$, as desired.

As for the last part, observe that the graph of $A(x)$ is a portion of a parabola that opens up. This will necessarily be maximized at one of its endpoints (if it opened down, it might maximize at its vertex, if in the domain). It is easy to check which of $x=20$ or $x=\frac{280}3$ maximizes $A(x)$, and what that maximum area is.

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Since the width of the rectangular field is $x$, the length is $2x$. Then the rectangular field requires $6x$ ft of fencing. This leaves $600 - 6x$ feet of fencing for the square field, so each side has length $150 - 1.5x$ ft.

The square field must contain at least 100 square ft, so $150 - 1.5x \ge 10$, implying that $x \le 93.333\ldots$.

The rectangular field must contain at least 800 square ft, so $2x^2 \ge 800$, implying that $x \ge 20$.

Thus, the domain for $x$ is $20 \le x \le 93.333\ldots$.

The total area contained in the two fields is $(150 - 1.5x)^2 + 2x^2 = 22500 - 450x + 4.25x^2$.

This is a quadratic function of $x$, and the parabola opens upward (as the leading coefficient is positive). Therefore, by the extreme value theorem, the maximum must be one of the endpoints of the domain.

If $x = 20$ then the area is $15200$. If $x = 93.3333\ldots$ then the area is $17522.2222\ldots$. Therefore the maximal area is $17522.2222\ldots$.

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  • $\begingroup$ Downvoting because I believe the above comment and because you haven't really provided any justification for the given "maximising" $x$-value to help me disbelieve it. I'll happily turn the downvote into an upvote for the rest of the question, if you fix it. $\endgroup$ – Ben Millwood Aug 9 '12 at 21:26
  • $\begingroup$ Oops, you guys are right - I worked through it too cursorily and made terrible assumptions. Edited. $\endgroup$ – khu Aug 9 '12 at 22:18

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