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WebGraphing.com Forum » List all forums » Forum: Calculus Homework Help » Thread: Maximum Area |
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Total posts in this thread: 3 |
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unluckykc
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Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 4. I know how to maximize, but I can't figure out an equation in order to maximize. |
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unluckykc
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I have to find it using different ways: a) using area as a function of the height, b) using area as a function of half the angle |
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pskinner
Joined: Apr 2, 2005 Posts: 797 Status: Offline |
Draw the circle with the inscribed isosceles triangle. From the center, draw the three radii to the vertices. Label the base b. Label the height of the smaller isosceles triangle with the same base whose vertex is the center of the circle, h. Using the formula for the area of a triangle, A=(1/2)b*height the height=h+r, where r is the radius, and b can be computed using the Pythagorean theorem, (r^2=(1/2)b^2+h^2), so you can get the Area as a function of the height alone. Differentiate with respect to the height to solve the problem this way. For the second way, label the angle of the bottom isosceles triangle theta. So, b=cos(theta) and h=rsin(theta), so the area is: A=(1/2)*b*(r+h). Substitute for b and h (r=4) and get A as a function of theta, so then you can solve the problem by differentiating. ---------------------------------------- Principal Skinner |
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Maximum Area
Re: Maximum Area