Optimizing Area in a Semi-Circle

James99x
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1. Find the dimensions of the rectangle with the largest area that can be inscribed in the upper semi-circle given by x^2+y^2 ≤ 16, y≥0.
2. I thought I'd use A=lw
3. This is but a guess..so take it with a grain of salt..

height=2x
base= x^2+y^2

A(x) = 2x(x^2+y^2)
= 2x^3+2xy^2

A'(x) = 6x^2+2y^2+4x(dy/dx)(y)


I'm not sure what to really do about this particular problem.
 
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Hint: start by drawing a diagram. You will see right away what you did wrong.

RGV
 
Are you familiar with LaGrange multipliers? If so, then this problem is easier than it seems.
 
Hi James99x! Welcome to PF! :smile:

You have the wrong "base" for your square...

Beyond that your method is fine! :wink:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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