- A truck rental company rents its 30 trucks by the day.?

Willetta Munhall: Rather than just give you the answer--that would be the easy way, but you wouldn't learn anything--I'm going to explain the process for solving a problem such as this. There are a few things I like to do when faced with a problem like this. First, identify the variable. Then, identify what exactly I'm being asked to solve. After identifying what the end result should be, I would then create the function(s) necessary to solve this problem.First off, let's identify the variable. If we were the company owner, what is it that we are changing in this scenario? We are changing the price. Specifically, we are changing the price by $1, either up or down. So, we can let the variable x="the number of $1 price increases". So, for example, when x=1, we are renting the trucks for $21/day. When x=-1, we are renting the trucks for $19/day.Now, in the problem, we are being asked to find the rental charge that maximizes the profit. Bells should be sounding in ! your mind right now, because this screams differentiation. We want to take the derivative of the profit function and find the value which maximizes this function. So, we need to create a profit function. Since profit is the difference of revenue and cost, we actually need to create a revenue function and a cost function. The cost, in this case, is very easy ($5/day, so C(x)=5). However, the revenue is a little more complicated. Revenue is price X quantity. We can create formulas for these values using the information we have been given.When we leave the price at $20/day, we rent all 30 trucks. When we rent for $21/day (x=1), we rent 29 trucks. When we rent for $22/day (x=2), we rent 28 trucks. Do you see the pattern?(price) p(x)=20+x(quantity) q(x)=30-xTherefore, revenue is R(x)=(20+x)(30-x)=600+10x-x^2.Before we continue, notice that negative x-values do not make sense in this instance, as we only have 30 trucks to rent (why rent the trucks for $19/day when we ca! n't rent 31 trucks?). In reality, your quantity function woul! d be piecewise, where q(x)=30 for xNow, remember that the profit is the difference of revenue and cost. So, profit is 600+10x-x^2 - 5, or P(x)=595+10x-x^2.Now that we have a profit function P(x), we must maximize it by taking the derivative, setting it equal to zero, and solving for x. P'(x)=10-2x. When we set this to zero and solve, we get x=5. ***WARNING*** Without verifying that this is the maximum, you cannot be sure. To do this, you can perform the 2nd derivative test: P''(x)=-2, so P''(5)=-2I hope this helped, and I hope that you can now figure out the remainder of your homework questions of this type. Best of luck!...Show more