2IntroductionIn this section we continue to solve optimization problems.We will see how to maximize a company’s profit if we are not given the price function, provided that we are given information describing how price changes will affect sales.We will also see that sometimes x should be chosen as something other than the quantity sold.
3Example 1 – FINDING PRICE AND QUANTITY FUNCTIONS A store can sell 20 bicycles per week at a price of $400 each. The manager estimates that for each $10 price reduction she can sell two more bicycles per week. The bicycles cost the store $200 each. If x stands for the number of $10 price reductions, express the price p and the quantity q as functions of x.Solution:Letx = the number of $10 price reductionsFor example, x = 4 means that the price is reduced by $40 (four $10 price reductions).
4Example 1 – Solutioncont’dTherefore, in general, if there are x $10 price reductions from the original $400 price, then the price p(x) isThe quantity sold q(x) will be
5Example 2 – MAXIMIZING PROFIT (Continuation of Example 1) Using the information in Example 1, find the price of the bicycles and the quantity that maximize profit. Also find the maximum profit.Solution:In Example 1 we foundp(x) = 400 – 10x Priceq(x) = x Quantity sold at that price
6Example 2 – Solution Revenue is price times quantity, p(x) · q(x): cont’dRevenue is price times quantity, p(x) · q(x):R(x) = (400 – 10x)(20 + 2x) p(x)q(x)= x – 20x Multiplying out and simplifyingThe cost function is unit cost times quantity:
7Example 2 – Solution Profit is revenue minus cost: cont’dProfit is revenue minus cost:We maximize profit by setting the derivative equal to zero:200 – 40x = Differentiating P = x – 20x2
8Example 2 – Solution The critical number is x = 5. cont’dThe critical number is x = 5.The second derivative, P"(x) = – 40, shows that the profit is maximized at x = 5.Since x = 5 is the number of $10 price reductions, the original price of $400 should be lowered by $ ($10 five times), from $400 to $350.The quantity sold is found from the quantity function:q(5) = · 5 = q(x) = x at x = 5
9Example 2 – Solution Finally, we state the answer clearly. cont’dFinally, we state the answer clearly.Sell the bicycles for $350 each.Quantity sold: 30 per week.Maximum profit: $ From P(x) = x – 20x2 at x = 5Although we did not need to graph the profit function, the diagram shown below does verify that the maximum occurs at x = 5.
10Choosing VariablesIn example 2 we did not choose x to be the quantity sold, but instead to be the number of $10 price reductions.We chose this x because from it we could easily calculate both the new price and the new quantity.Other choices for x are also possible, but in situations where a price change will make one quantity rise and another fall, it is often easiest to choose x to be the number of such changes.
11Example 3 – MAXIMIZING HARVEST SIZE An orange grower finds that if he plants 80 orange trees per acre, each tree will yield 60 bushels of oranges. He estimates that for each additional tree that he plants per acre, the yield of each tree will decrease by 2 bushels. How many trees should he plant per acre to maximize his harvest?Solution:We take x equal to the number of “changes”—that is, letx = the number of added trees per acre
12Example 3 – Solution With x extra trees per acre, cont’dWith x extra trees per acre,Trees per acre: x Original 80 plus x moreYield per tree: – 2x Original yield less 2 per extra treeTherefore, the total yield per acre will beWe maximize this by setting the derivative equal to zero:–100 – 4x = Differentiating Y = 4800 – 100x – 2x2x = – Negative!
13Example 3 – Solutioncont’dThe number of added trees is negative, meaning that the grower should plant 25 fewer trees per acre.The second derivative, Y"(x) = –4,shows that the yield is indeedmaximized at x = –25.Therefore:Plant 55 trees per acre – 25 = 55
14Example 4 – MINIMIZING PACKAGE MATERIALS A moving company wishes to design an open-top box with a square base whose volume is exactly 32 cubic feet. Find the dimensions of the box requiring the least amount of materials.Solution:The base is square, so we definex = length of side of basey = height
15Example 4 – Solution A = x2 + 4xy cont’dThe volume (length · width · height) is x · x · y or x2y, which (according to the problem) must equal 32 cubic feet:x2y = 32The box consists of a bottom (areax2) and four sides (each of area xy).Minimizing the amount of materials means minimizing the surface area of the bottom and four sides:A = x2 + 4xy
16Example 4 – Solutioncont’dAs usual, we must express this area in terms of just one variable, so we use the volume requirement to express y in terms of x:The area function becomes
17Example 4 – Solution We minimize this by finding the critical number: cont’dWe minimize this by finding the critical number:
18Example 4 – Solution The second derivative cont’dThe second derivativeis positive at x = 4, so the area is minimized.Therefore, the dimensions using the least materials are:Base: 4 feet on each side Height from y = 32/x2 at x = 4Height: 2 feet
19Minimizing the Cost of Materials How would the preceding problem have changed if thematerial for the bottom of the box had been more costlythan the material for the sides?If, for example, the material for the sides cost $2 persquare foot and the material for the base, needing greaterstrength, cost $4 per square foot, then instead of simplyminimizing the surface area, we would minimize total cost:
20Minimizing the Cost of Materials Since the areas would be just as before, this cost would beCost = (x2)(4) + (4xy)(2) = 4x2 + 8xyFrom here on we would proceed just as before, eliminatingthe y (using the volume relationship x2y = 32) and thensetting the derivative equal to zero.
21Maximizing Tax Revenue Governments raise money by collecting taxes. If a sales tax or an import tax is too high, trade will be discouraged and tax revenues will fall.If, the tax rate is too low, trade may flourish but tax revenues will again fall.Economists often want to determine the tax rate that maximizes revenue for the government. To do this, they must first predict the relationship between the tax on an item and the total sales of the item.
22Maximizing Tax Revenue Suppose, for example, that the relationship between the tax rate t on an item and its total sales S isIf the tax rate is t = 0 (0%), then the total sales will beIf the tax rate is raised to t = (16%), then sales will be
23Maximizing Tax Revenue That is, raising the tax rate from 0% to 16% will discourage $8 million worth of sales.The graph of S(t) as shows how total sales decrease as the tax rate increases.With such information (which may be found from historical data), one can find the tax rate that maximizes revenue.
24Example 5 – MAXIMIZING TAX REVENUE Economists estimate that the relationship between the tax rate t on an item and the total sales S of that item (in millions of dollars) isFind the tax rate that maximizes revenue to the government.Solution:The government’s revenue R is the tax rate t times the total sales :
25Example 5 – Solutioncont’dTo maximize this function, we set its derivative equal to zero:
26Example 5 – Solution This gives a tax rate of t = 9%. cont’dThis gives a tax rate of t = 9%.The second derivative,is negative at t = 0.09, showing that the revenue is maximized.Therefore:A tax rate of 9% maximizes revenue for the government. See the graph as shown.