Question

A rectangle is constructed with its base on the x-axis and two of its vertices on the parabola y = 16 - x? What are the dimensions of the rectangle with the maximum area? What is the area?The shorter dimension of the rectangle is and the longer dimension is I(Round to two decimal places as needed.)

Answer 1

A = l * w

We know the w is 2 times the x distance and the l is the y distance

A = 2x * y

A = 2x * ( 16-x^2)

A = 32x - 2x^3

Take the derivative to find the maximum and then set that equal to zero to find the x value

0 = d/dx ( 32x - 2x^3)

0 = 32 - 6x^2

32 = 6x^2

32/6 = x^2

Taking the square root of each side

sqrt(16/3) = x

4 sqrt(3)/3 =x

2x= 8 sqrt(3)/3

Now find y

y = 16 - x^2 = 16 - 32/6 = 16 - 16/3 = 48/3 - 16/3 = 32/3

The final step is find the area

A = 2 xy = 2 * 4 sqrt(3)/3 * 32/3

=256 sqrt(3)/9 which is approximately 49.27