Tuesday

March 3, 2015

March 3, 2015

Posted by **Stuck** on Tuesday, March 9, 2010 at 10:25pm.

There is a diagram, but I think the question makes it clear enough what is going on. I'm having problems finding a relationship that I can work with.

- Calculus -
**Reiny**, Tuesday, March 9, 2010 at 10:45pmdraw a radius from the centre to a vertex of the rectangle

Let the base of the rectangle be 2x, making the triangle base x, and let the height of the rectangle by y

Area of rectangle = 2xy

but x^2 + y^2 = 4

y = √(4 - x^2) or (4-x^2)^(1/2)

A = 2x(4-x^2)^(1/2)

take the derivative using the product rule,

set it equal to zero and solve for x

- Calculus -
**Stuck**, Tuesday, March 9, 2010 at 11:21pmThank you! Never occurred to me to position the radius like that.

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