Solve this derivative related problem.?

Question

A rectangel has two of its corners on the curve y = 16 - x^2, - 4 <= x <=4 and the othe corners on the x-axis. Find the maximam possible are of this rectangle.

Answer 1

If you have a corner on 16 - x^2, then your width is 2x and your height is simply 16 - x^2. the area then is 2x(16 - x^2), or 32x - 2x^3.

Finding the max area involves a derivative, so f'(x) = 32 - 6x^2. From here, we set this to 0 and solve for x:

0 = 32 - 6x^2
0 = 16 - 3x^2
16 = 3x^2
16/3 = x^2
x = sqrt(16/3)
x = 4sqrt(3)/3

So for this x, we will have the maximum area. If we plug this in to our original equation, we find that the area is:

f(x) = 32x - 2x^3
f(4sqrt(3)/3) = 32(4sqrt(3)/3) - 2(4sqrt(3)/3)^3
= 128sqrt(3)/3 - 64sqrt(3)/9
= 320sqrt(3)/9
~ 61.58

Just as a check, take a look at what happens aorund x. We found x to be 4sqrt(3)/3, or about 2.31. If we used x = 2 in our area formula, we get:

f(2) = 32(2) - 2(2)^3
= 64 - 16
= 48

And for something larger than 2.31, like 2.5, we get:

f(2.5) = 32(2.5) - 2(2.5)^3
= 80 - 31.25
= 48.75

We can see that the areas get smaller as you go away from 4sqrt(3)/3.

So, the max area you'll get is 320sqrt(3)/9, or about 61.58.

Answer 2

is that this a calculus undertaking? would not appear as if it. First calculate the circumference of the ferris wheel which would be 2 x pi x radius or 2 x 3.142 x 6 = 37.704m If its rotating at a million revolution in conserving with 2 minutes any element on the wheel strikes at 18.852m/min or 0.3142m/sec or a million.131km/hr If a rider is 8m above the floor he maintains to be moving at a million.131km/hr do you may make sure the horizontal and vertical velocities? if so you will could desire to word algebra.