To create the box, you are asked to cut out smaller squares from each corner of the 8 by 8 inches piece of cardboard and to fold up the flaps to form the sides.
According to U.S.Β postal regulations, the girth plus the length of a parcel sent by mail may not exceed 108 inches, where the βgirthβ is the perimeter of the smallest end. What is the largest possible volume of a rectangular parcel with a square end that can be sent by mail? What are the dimensions of the package of largest volume?
Let \(x\) represent the length of one side of the square end and \(y\) the length of the longer side. Label these quantities appropriately on the image shown in FigureΒ 74.
The problem statement tells us that the parcelβs girth plus length (\(P\)) may not exceed 108 inches. In order to maximize volume, we assume that we will actually need the girth plus length \(P\) to equal 108 inches. What equation does this constraint give us involving \(x\) and \(y\text{?}\)
The equation above gives the relationship between \(x\) and \(y\text{.}\) For ease of notation, solve this equation for \(y\) as a function on \(x\) and then find a formula for the volume of the parcel as a function of the single variable \(x\text{.}\) What is the formula for \(V(x)\text{?}\)
Over what domain should we consider this function? To answer this question, notice that the problem gives us the constraint that \(P\) (girth plus length) is 108 inches. This constraint produces intervals of possible values for \(x\) and \(y\text{.}\)
Use calculus to find the global maximum of the volume of the parcel on the domain you just determined. Justify that you have found the global maximum using either the Closed Interval Method, the First Derivative Test, or the Second Derivative Test!
Notice that a critical point might or might not be a global maximum or minimum, so just finding the critical points is not enough to answer an optimization problem. Moreover, some of the critical points might be outside of the domain imposed by the context and thus they cannot be feasible optimal points.
Activity3.8.5.Revenue = Number of tickets \(\times\) Price of ticket.
Waterford movie theater currently charges $8 for a ticket. At this price, the theater sells 200 tickets daily. The general manager wonders if they can generate more revenue by increasing the price of a tickets. A survey shows that they will lose 20 customers for every dollar increase in the ticket price.
Consider the new price of a ticket \(P(d)\) and the new number of tickets sold \(T(d)\text{.}\) Write a formula for the revenue earned by ticket sales \(R(d)\) as a function of a price increase of \(d\) dollars.
What increase in price \(d\) should the general manager choose to maximize the revenue? What price would a movie ticket cost then and what would the revenue be at that price?
Suppose now that the cost of running the business when the price is increased by \(d\) dollars is given by \(C(d) = 10d^3 β40d^2 +40d+600\text{.}\) If the manager decides that they will definitely increase the price, what price increase \(d\) maximizes the profit? (Recall that Profit = Revenue - Cost).
The city council is planning to construct a new sports ground in the shape of a rectangle with semicircular ends. A running track 400 meters long is to go around the perimeter.
Suppose that if a widget is priced at \(\$176\text{,}\) then you are able to sell \(672\) units each day. According to a survey of customers, increasing this price by \(\$1\) will result in losing \(4\) daily sales; decreasing by \(\$1\) will gain \(4\) daily sales. Your manager asks you how to adjust the price of a widget to maximize the revenue (widgets sold times price). Write an explanation of what this change in price should be and why.