The third midterm is here.

Solutions follow.

**Problem 1.** *Consider the region bounded below by the curve and above by the -axis. A rectangle is inscribed in this region in such a way that one of its sides lies on the -axis and the other two corners are on the parabola. Find the dimensions of the sides of the rectangle that maximize its area.*

The graph above (click to enlarge) shows the curve together with both axes, and a rectangle as required. I have marked the coordinates of one of the corners of the rectangle, as .

Suppose that and are the dimensions of the sides of the rectangle, with the length of the horizontal sides. Then the area of the rectangle is , and in order to apply calculus to maximize this function, we first express and in terms of .

We have that and . Hence, if denotes the area of the rectangle, as a function of , we have that

Note that, by our convention, . We have , so , or , or

or

Since , it must be the case that . Since , it follows that this value of gives us a local maximum of the function , and therefore the only maximum in the interval , which is the range of values that can take and be a length of a rectangle as required.

It follows that

and ,

and the maximum area is .

**Problem 2.** *Two (very long) roads cross each other at a point . One runs north to south, and the other runs east to west. Car is driving north along the first road, and car is driving east along the second road. At a particular time, car is 30 miles to the north of and traveling at 50 m/h, while car is 40 miles to the east of and traveling at 65 m/h. How fast is the distance between the two cars changing? *

The graph above (click to enlarge) shows the situation described in the problem. Given an arbitrary time , let’s denote by the distance from to the point of intersection of the roads, by the distance from to , and by the distance between and , all computed at time .

The problem is asking us to compute at the time that makes and , where units are measured in miles, and time is measured in hours.

Note first that , by the Pythagorean theorem. Then . or

At the time that we are interested in, we have , , , and , so

m/h.

**Problem 3.** *Let . Find a point between and with the property that coincides with the slope between the endpoints of on .*

The slope of the line between the endpoints of on is simply

,

so we need to solve the equation

.

For this, note that , so and we have

, or , or .

In the graph above (click to enlarge) I have displayed in different colors the line and the line tangent to at . I have also shown the line and the curve . Note how the tangent line is indeed parallel to the given secant line.

**Problem 4.** *The equation obviously has solution . Show the values of obtained by applying Newton’s method to this equation, starting with .*

Newton’s method begins with a guess (in this case ) and iteratively improves it, approaching a solution of the equation In this case, . The iterative process is given by the equations

for

In our case, , so and

,

so , i.e., in our iterative process we have

for

This means that:

- ,
- ,
- , and
- .

The graph above (click to enlarge) shows the function , and the “staircase” produced by applying Newton’s method starting with and ending with .

**Problem 5.** *Find all functions such that *

* *

*and .*

From results in class, we know that if for all in the common domain of and then (on each interval where both and are defined) there is a constant such that .

What this means for the problem at task is that it suffices to guess a function with the given derivative, because then any other function with the same derivative will simply differ from our guess by a constant. From the derivatives studied through the course, we easily see now that

.

To find , we use that : From the formula above, we have that

or , so

.

The two graphs below (click to enlarge) show first and then , with a little circle indicating the position of the point on the graph.