Maximizing Triangle Area with Calculus

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Homework Statement



The vertices of a triangle are (0,0), (x, cos x), and (sin3x, 0), where 0 < x < p/2.
a. If A(x) represents the area of the triangle, write a formula for A(x).
b. Find the value of x for which A(x) is a maximum. Justify your answer.
c. What is the maximum area of the triangle?

Homework Equations


I know this involves min/max problems, and the area of a triangle at some point.

The Attempt at a Solution


No idea where to start. I don't see any way to get an area function for the triangle.
 
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You do know that the area of a triangle if "1/2 base times height", don't you? Draw a picture and see what base and height are for this triangle.
 
So is A(x)= (1/2)(sinx)^3(cosx)? If that's true then do I have to take the derivative or something?
 
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Is it sin3x or (sinx)^3? Once that is cleared up, how would you usually find the maximum of a function? A derivative would help.
 
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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