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Refer to the graph below:

The points $(3,0)$, $(x,0)$, $(x, \frac{1}{x^2}),$ and $(3, \frac{1}{x^2})$ are the vertices of a rectangle where x $\ge$ 3, as shown in the figure below. For what value of x does the rectangle have a maximum area?

(A) $3$

(B) $4$

(C) $6$

(D) $9$

(E) there is no such value of x

I know the answer is (C), but I don't understand the steps to solving this problem.

If anyone could tell me the steps to solve this question and why each step is necessary, that would be super helpful!

Thank you.

enter image description here

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    $\begingroup$ Write the area in terms of the four vertices. This area will turn out to be a function of $x$. Now maximize this function w.r.t. $x$. $\endgroup$ – avs Feb 3 '17 at 0:17
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    $\begingroup$ Or write area as base times height, base is $x-3$ and height $1/x^2$. $\endgroup$ – coffeemath Feb 3 '17 at 0:35
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The rectangle has sides $x-3$ and $\frac{1}{x^2}$, so the area is:

$$f(x)=\frac{x-3}{x^2}$$

And: $$f'(x)=\frac{6-x}{x^3}$$ which is positive for $x<6$ ($f$ increasing) and negative for $x>6$ ($f$ decreasing)

So the max value is attained for $x=6$

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