is 2r + 2h + πr = 288. We can use the equation for the perimeter to eliminate a variable from
the equation for the area. Let’s isolate h: h = 144 − r − . Now we can substitute for h in the
equation for the area: A = = . Next, we can take
the derivative: = 288 − 4r − πr. If we set this equal to zero and solve for r, we get r =
inches. We can verify that this is a maximum by taking the second derivative: = −4
− π. We can see that the value of the second derivative is negative. Therefore, according to the
second derivative test (see this page), the area is a maximum at r = inches.
- θ = radians (or 45 degrees)
We simply take the derivative: (2cos 2θ). Note that v 0 and g are constants, so the
only variable we need to take the derivative with respect to is θ. Now we set the derivative
equal to zero: (2cos 2θ) = 0. Although this has an infinite number of solutions, we are
interested only in values of θ between 0 and radians (why?). The value of θ that makes the
derivative zero is θ = radians (or 45 degrees). We can verify that this is a maximum by
taking the second derivative: = (−4sin 2θ). We can see that the value of the second
derivative is negative at θ = . Therefore, according to the second derivative test (see this
page), the range is a maximum at θ = radians.
- 10 feet by 10 feet
Let’s label the sides of the box. The base’s length and width are the same (because it’s a
square) so let’s call them both x. Let’s call the height of the box y. This makes the volume of the
box V = x^2 y. The surface area is the area of the bottom, x^2 , plus the areas of the four sides, each
of which is xy. There is no top, so the surface area is A = x^2 + 4xy.
We are given that the volume is 500 ft^3 , so x^2 y = 500. Let’s isolate y, and then plug the
