Maximum area?

Question

A rectangle is bounded by the x-axis and the semicircle y=sqrt(25 - x^2). What length and width should the rectangle have so that its area is a maximum?

Answer 1

Note that choosing one corner on the semicircle defines the rectangle. Suppose we choose the corner at
(x, sqrt(25-x^2) ), where x is positive (and less than 5).

Then it is easy to see the width is 2x, and the height is
sqrt (25-x^2)
so the formula for the area is
2x * sqrt(25 -x^2)
Now use calculus to find the value of x that maximizes this value, and plug it in to get the maximum area.

HTH,
Steve

Answer:
x = 5/Sqrt(2), area = 25

EDIT - I'm not entirely sure how everyone else is getting this wrong, but the area is
2 * x * sqrt(25-x^2)
not 4, not 1, but 2. We are talking about a semicircle!

Answer 2

Area of a rectangle, A = xy
dA/dx = d/dx (xy) = y + x dy/dx =set= 0 to find the extrema
dy/dx = (1/2)(25-x^2)^(-1/2) d/dx(25-x^2)
= (1/2)(25-x^2)^(-1/2)(-2x)
= -x(25-x^2)^(-1/2)
Now put this back into the eqn for dA/dx
(25-x^2)^(1/2) - x^2(25-x^2)^(-1/2) = 0
(25-x^2) - x^2 = 0 by multiplying both sides by (25-x^2)^(1/2)
2x^2 = 25
x = 5/sqr[2]
Now just sub x back into the eqn for y:
y = sqr[25 - 25/2] = 5/sqr[2]

Answer 3

Hello,

The area is 4*x*(sqrt(25-x^2))

Now to get the max we need to take the derivative so we have.

a' = 4*(sqrt(25-x^2))+.5*sqrt(25-x^2)^(-.5)*(-2x)*(4x)= 4*sqrt(25-x^2)-x*sqrt(25-x^2)^(-.5)*4x
Now set this = 0 and multiply by sqrt(25-x^2) and we get 0 = 4*(25-x^2) - 4x^2=0

I'll let you finish.

Hope This Helps!!