A rectangle is bounded by the x-axis and the semicircle y=�ã(25-x^2).

Question

Question is, what length and width should the rectangle have so that its area is a maximum, and what is the maxuimum area?


Area= length*width
= 2x*y= 2x*sqrt(25-x^2)

Now, take that, differentiate it, set to zero, and solve for x,y. Length = 2x, width (or height) is y.
I will be happy to critique your work or thinking.

Answer 1

Differentiating with respect to x:

2*sqrt(25-x^2) - 4x*(1/2)*(25-x^2)^(-1/2) = 0

2*sqrt(25-x^2) = 4x*(1/2)*(25-x^2)^(-1/2)

4x^2 = 25-x^2

5x^2 = 25

x = sqrt(25/5)

y = sqrt(25-x^2)

= sqrt(25-(25/5))

= sqrt(20/5)

Therefore, the length and width of the rectangle should be 2*sqrt(25/5) and sqrt(20/5) respectively, and the maximum area is (2*sqrt(25/5))*(sqrt(20/5)) = 4*sqrt(500/25) = 4*sqrt(20) units^2.