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find the area of the largest rectangle that can fit inside a semi circle of radius 2 cm I have absolutely no idea where to get started on this...What I did do is $A=(\pi r^2)/2$ (its a semi circle)

that gave me $6.28\ \mathrm{cm}^2$ now what do I do?

I

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Let's agree that we only have to consider rectangles with two vertices on the diameter of the semicircle. This means that the four vertices have coordinates $\pm(r\cos\phi,0)$ and $\pm(r\cos\phi, r\sin\phi)$ for some $\phi\in\bigl[0,{\pi\over2}\bigr]$. The area of such a rectangle is $2r^2\cos\phi\sin\phi=r^2 \sin(2\phi)$, and this is maximal when $\phi={\pi\over4}$. Therefore the maximal possible area of such a rectangle is $r^2$.

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Hint: Imagine the semicircle being above the $x$ axis, so it is $y=\sqrt{4-x^2}$. If you are given the width (which lies along $x$), can you calculate the height and then the area? Then take the derivative of the area with respect to width, set to zero, etc.

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This is a pretty straightforward optimization problem. Here's how we want to go about it:

  1. Write down constraints and the function to optimize
  2. Use constraints to express the function in terms of one variable (usually using substitution)
  3. Take the derivative of the function to optimize and set it equal to zero

The value we get when we set the derivative equal to zero is the optimal value for that variable, and the values of other variables can be found from the expressions we have in step 1.

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