# Mathematics - Maximum Area

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An isosceles triangle has two equal sides of length 10cm. Let theta be the angle between the two equal sides. The Area of the triangle can be expressed as: A(theta) = 50sin(theta)

If theta is increasing at the rate of 10 degrees per minute (pie/18 radians per minute)and the derivative of the area is [50cos(theta)]at what value of theta will the triangle have a maximum area?

I thought:

Because [50cos(theta)]=0 when theta= (pie/2)

and because the second derivative of the area is less than zero, we know there is an absolute maximum when theta=(pie/2)

And that the maximum area could be found by placing (pie/2) into the original area equation A(theta)= 50sin(pie/2)=50 cm^2

However, when a theta is (pie/2) doesn't that mean that it is no longer a triangle?

How would you answer this question?

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A = (1/2) ab sin C

A = (1/2)(10 * 10) sin theta

A = 50 sin theta

dA/dtheta = 50 cos theta

For maximum area, dA/dtheta ...

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