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A rectangle is inscribed with its base on the $x$-axis and its upper corners on the parabola $y=7−x^2$. What are the dimensions of such a rectangle with the greatest possible area?

Would this be a basic optimization problem, with the constraint being the area of the rectangle $ A = bh $ and the objective function being the derivative of $y = 7 - x^2$. Just a little confused on how to get this started.

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    $\begingroup$ You've got it backwards. The objective is the area. The constraint is that the upper corners are on the parabola. $\endgroup$ – Michael Grant Mar 27 '15 at 2:54
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Note that you can easily see how the rectangle is defined simply from one value of $y$. So express the area of the rectangle in terms of that $y$ value and then you can optimize that expression.

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Hint

Consider that the lower corners have coodinates $(-x,0)$ and $(+x,0)$. Since the upper corners are along the parabola, their coordinates are $(-x,7-x^2)$ and $(+x,7-x^2)$. So, the area of the rectangle is $$A(x)=2x(7-x^2)$$ and this is the quantity you want to maximize.

What does imply the fact that a function goes through an extremum ? Just apply and when you get the solution, do not forget to apply a famous test which tells if the extremum is a maximum or a minimum.

I am sure that you can take from here.

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Let the height of the rectangle be $h$. Then the base of the rectangle are the points that solve $h=7-x^2$. That is: $\pm\sqrt{7-h}$ and so has a length of $2\sqrt{7-h}$. So, $A=2h\sqrt{7-h}$. Optimise.

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Since the function $f$ is symmetric, $A(x)=2xf(x)$.

You need to consider the derivative of $A$ for $x\in[0,x_{\max}]$, where $f(x_{\max})=0$, $x_{\max}>0$.

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