CHAPTER 2 SUPPLEMENTARY EXERCISES
6. Polynomials of degree n have at most n − 1 “turns,” are differentiable (smooth and unbroken), and tend to infinity at extremes
3.5 MARGINAL ANALYSIS
MARGINAL ANALYSIS 103 19. What is the maximum area of a rectangular garden that can be fenced for $120 if fencing on three sides of the garden cost $5 per linear foot and on the fourth side at $7 per linear foot?
20. A Norman Window is a rectangular window capped by a semicircular arch. Find the value ofrso the window perimeter is 20 feet and has maximum area. Lethrepresent the height, andrthe radius of the semicircle. Therefore, the width will be 2r.
21. Suppose that three sides of a rectangular pen to be built with material that costs $10 per linear foot and a less visible fourth side uses a material that costs $5 per linear foot.
If $2400 is available to build the pen, what are the dimensions for largest area?
22. An enclosure of 288 square yard area is to be built. Three sides of the enclosure cost
$16 per running yard, while the more visible fourth side costs $20 per running yard.
Find the dimensions of the enclosure that will minimize the cost of building and state what this minimum is.
23. A closed rectangular box with a square base is to be formed using 60 square feet of material. What dimensions maximize volume?
24. Show that the optimal length for a mailing container of maximum volume with the length plus girth restriction of 108′′is 36′′when the cross-section is
(a) an equilateral triangle. (b) an arbitrary cross-section geometry.
Hint: Assume that the cross-sectional areaA, is proportional to a dimensionx, so A=kx2, and the perimeter,p, is proportional tox, say,p=bxwherekandbare pro- portionality constants.
25. Suppose that in Example 3.4.8 the circular ends are cut from a large sheet that has been divided into squares of side 2r. Obtain the optimal height to diameter ratio of the mailing tube.
26. To minimize the waste in cutting circular ends from square pieces, a suggestion is made to divide the large sheet into hexagons (as a honeycomb). Show that the optimal height to diameter ratio becomes2√
3
𝜋 ≈1.1.
Hint: The area of a regular hexagon is
√3
2 W2whereWis its width and can diameter.
27. A vertical cylindrical tank of given volume is to be constructed. The required material for the top and bottom costs twice as much per square foot as that for the sides. Find the best height to diameter ratio.
28. A strip of lengthLis to be cut into 12 smaller strips to form a parallelepiped with square base and of maximum volume. How shall the strip be divided?
as price increases. Calculus aids in answering useful questions for a functional relation between price,p, and demand,x, (assumed to be a continuous and differentiable).
Economists define several functions of demand,x, such as unit price of an item as a func- tion of demand,p(x); total revenue,R(x); and total cost,C(x); among others. The derivatives of these functions, being rates of change, are theirmarginals. For example, themarginal cost (MC)is the derivative of the cost function,C′(x). Economists interpret MC as thecost of an additional unit of demand. This is distinct fromaverage cost,which is the total cost divided by the number of items,C(x)/x.
Similarly, as revenue is the unit price multiplied by demand, a total revenue function is R(x) =xp(x).Marginal revenue (MR)is the derivative,R′(x). It is thechange in revenue accompanying a unit change in demand.
It is important to understand why economists use the derivative as the MC. Imagine a small change,Δx, in demand,x, fromxtox+ Δx. Then, the total cost changes fromC(x) toC(x+ Δx). The change in cost per unit change,Δx, in demand is
C(x+ Δx) −C(x) Δx
You may recognize (from Chapter 2) this as the difference quotient whose limit as Δx→0 is the derivative of C(x). Understandably, calculus has a vital role in such studies.
Example 3.5.1 Maximizing Revenue
The price of an item as a function of demand, x, is p(x) =5−x∕4 dollars. What production level maximizes revenue? What is the maximum revenue?
Solution:
Total revenue, demand multiplied by unit price, is R(x) = (x)(5−x∕4) =5x−x2∕4.
Maximizing revenue implies vanishing marginal revenue, R′(x) =0.
Therefore, setting R′(x) =0=5−x∕2 yields x=10.
The derivative is positive for x<10 and negative for x>10 so at x=10, R(10) =25.
A demand of 10 units maximizes revenue at $25.
Example 3.5.2 Marginal Cost
Obtain and interpret the MC for the cost function C(x) =x3–12x2+36x+10 and a demand of x units.
Solution:
The derivative of the cost function is the MC, so
MC(x) =C′(x) =3x2−24x+36
It is the change in the total cost for a unit change in demand at level x.
MARGINAL ANALYSIS 105
Example 3.5.3 Marginal Revenue and Marginal Cost
For R(x) = −x2
2 +150x and C(x) = x3
30−2x2+110x+1600 determine the marginal rev- enue and marginal cost functions.
Solution:
The derivative of the revenue function is the marginal revenue, so MR(x) = −x+150
The derivative of the cost function is the marginal cost, so MC(x) =C′(x) = x2
10−4x+110
It is the change in the total cost for a unit change in demand when demand is at level x.
Note: we will revisit this again in Chapter 6. There, using marginal functions and initial conditions, help obtain revenue and cost functions.
Example 3.5.4 Minimizing Marginal Cost
For C(x) =2x3−12x2+40x+60, determine minimum marginal cost.
Solution:
Firstly, obtain MC(x), the derivative of C(x) as MC(x) =C′(x) =6x2−24x+40. For a minimum, set the derivative of MC(x) to zero. That is,
MC′(x) =C′′(x) =12x−24=0
When x<2, the derivative of MC is negative, and when x>2, its derivative is positive.
This indicates a minimum at x=2. The minimum marginal cost is MC(2) =6(2)2−24(2) +40=$16.
(A common mistake uses C(2) instead of MC(2)!)
Example 3.5.5 Maximizing Profit Demand, x, for an item at price, p, D(p), is
D(p) =x=50− (p∕3) What production level and price maximizes profit?
The cost function is
C(x) = (2∕3)x3+2x2−150x+1200
Solution:
To start, investigate restrictions on values of x and p, as there could be endpoint extrema.
Increasing price decreases demand, and conversely.
Therefore, seek intercepts in the first quadrant.
From the demand function, D(p), conclude that demand cannot exceed 50 units (when p=0) and price cannot exceed $150 (or demand becomes negative). That is,
0≤x≤50 and 0≤p≤150
Next, to obtain R(x), solve for price as a function of demand, x. Therefore, p(x) =150−3x and R(x) =x p(x) =150x−3x2 Subtracting cost from revenue yields the profit function Pr(x),
Pr(x) =R(x) −C(x) = (150x−3x2) − ((2∕3)x3+2x2−150x+1200)
= (−2∕3)x3−5x2+300x−1200 To maximize the profit, set its derivative to zero as
Pr′(x) = −2x2−10x+30=0
= −2(x+15)(x−10) =0
While there are two mathematical possibilities: either x= −15 or x=10, only x=10 sat- isfies the constraint x≤50 so Pr(10) =$633.33; a possible maximum.
However, there may be endpoint extrema. Evaluating Pr(x) at x=0 and x=50 yields:
Pr(0) = −$1200 Pr(50) = −$82,033.33
Clearly, these are not usable. There are no endpoint extrema of concern here.
The optimal option is to produce 10 units for sale at a unit price of $120(=150−3(10)) for a maximum profit of $633.33.
EXERCISES 3.5
1. What is the marginal cost forC(x) =2x3+9x+25?
2. What is the marginal cost forC(x) =x5−5x2+20?
3. IfC(x) =x7−5x3+20x+25, what is the marginal cost function whenx=1? (hint:
findC′(1))
4. When marginal cost will be zero forC(x) =2x4−36x2+70?
5. When marginal cost will be zero forC(x) =x3−9x2+15x+120?
6. Determine the minimal marginal cost forC(x) =4x3−24x2+50x+28.
MARGINAL ANALYSIS 107
7. Determine the minimal marginal cost forC(x) =x3−12x2+60x+20.
8. When the revenue function isR(x) =200√
x, what is the marginal revenue function, R′(x)?
9. When the revenue function isR(x) = −x2+30x, what is the marginal revenue function, R′(x)?
10. If price isp=6− (1∕2)x, determine the revenue and marginal revenue functions.
11. Ifp(x) = (−1∕4)x+12, find the revenue and marginal revenue functions.
12. Consumer demand for an item as a function of its pricepisx=D(p) =50− (p∕3).
Determine the production level and price that maximizes profit when the cost function isC(x) = (2∕3)x3+2x2−150x+1000.
13. Consumer demand for an item as a function of its pricepisx=D(p) =40− (p∕6).
Determine the production level and price that maximizes profit if the cost function isC(x) =x3+9x2−360x+2000.