SOLUTION: A rectangle has one vertex in quadrant 1 on the graph of y=16-x^2, another at the origin, one on the positive x-axis, and one on the positive y-axis. Express the area A of the r

Algebra ->  Surface-area -> SOLUTION: A rectangle has one vertex in quadrant 1 on the graph of y=16-x^2, another at the origin, one on the positive x-axis, and one on the positive y-axis. Express the area A of the r      Log On


   



Question 954959: A rectangle has one vertex in quadrant 1 on the graph of y=16-x^2, another at the origin, one on the positive x-axis, and one on the positive y-axis.
Express the area A of the rectangle as a function of x.
My answer: A(x)= 16x-x^3
Find the largest area A that can be enclosed by the rectangle?

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
The area of the rectangle would be,
A=x%2Ay
A=x%2816-x%5E2%29
A=16x-x%5E3
So far you are correct.
.
.
.
To find the maximum, take the derivative with respect to x and solve when the derivative equals zero.
dA%2Fdx=16-3x%5E2=0
3x%5E2=16
x%5E2=16%2F3
x=4%2Fsqrt%283%29
x=%284%2F3%29sqrt%283%29
So then,
A%5Bmax%5D=16%284%2F3%29sqrt%283%29-%2816%2F3%29%284%2F3%29sqrt%283%29
A%5Bmax%5D=%28128%2F9%29sqrt%283%29