□ 3.7.1 Guideline.
We follow the guideline to solve an optimization problem.
1. If there’s a picture to draw, draw it! Don’t try to visualize how things look in your head. Put a picture down on paper and label it.
2. Determine what the variables are and how they are related.
3. Decide what quantity needs to be maximized or minimized.
4. Write an expression for the quantity to be maximized or minimized in terms of only one variable. To do this, you may need to solve for any other variables in terms of this one variable.
5. Determine the minimum and maximum allowable values (if any) of the variable you’re using.
6. Solve the problem. (Be sure to answer the question that is asked.)
□ 3.7.2 Area.
Figure 3.2: Rectangular Space
Example 3.7.1. You have 40 (linear) feet of fencing with which to enclose a rectangular space for a garden.
Find the largest area that can be enclosed with this much fencing and the dimensions of the corresponding garden. See the figure3.2.
ANSWER. Let x and y be width and length of the rectangle, respectively. Then the area A is expressed by
A = xy. (3.7.1)
The given information implies
2x + 2y = 40 −→ x + y = 20. (3.7.2)
That is, we want to find the maximum value of A = xy where x and y satisfies x + y = 20.
From the equation (3.7.2), we have y = 20− x. Putting it into the equation (3.7.1), we get A(x) = x(20− x).
The maximum value of A(x) = x(20− x) occurs at the critical number x = 10, which implies y = 10. Thus
the largest area is A = 100. □
Example 3.7.2. You have 8 (linear) feet of fencing with which to enclose a rectangular space for a garden.
Find the largest area that can be enclosed with this much fencing and the dimensions of the corresponding garden.
ANSWER. The largest area is A = 4 with width 2 and length 2. □ Example 3.7.3. Find the area of the largest rectangle that can be inscribed in the ellipse x2
a2+y2
b2 = 1, where a and b are nonzero constants.
Figure 3.3: Cardboard
□ 3.7.3 Volume.
Example 3.7.4. A square sheet of cardboard 18′′ on a side is made into an open box (i.e., there’s no top), by cutting squares of equal size out of each corner and folding up the sides along the dotted lines. Find the dimensions of the box with the maximum volume. See the figure3.3.
ANSWER. Let x be the length of one side of the square cut. Then the length of the side of the square sheet becomes 18− 2x. So the volume V of the box becomes V(x) = x(18 − 2x)2, because it has the width 18− 2x and length 18− 2x and height x. We want to maximize the volume V. From the graph of x(18 − 2x)2 or by the argument on critical numbers and the maximum values, we deduce the maximum value of the volume occurs at the critical number x = 3. Thus the maximum volume is V = 432 with width 12, length 12, and
height 3. □
Example 3.7.5. A square sheet of cardboard 6′′ on a side is made into an open box (i.e., there’s no top), by cutting squares of equal size out of each corner and folding up the sides along the dotted lines. Find the dimensions of the box with the maximum volume.
ANSWER. The maximum volume is V = 16 with width 4, length 4 and height 1. □ Example 3.7.6. If 1200 cm2of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
□ 3.7.4 Closest Point to Curve.
Example 3.7.7. Find the point on the parabola y = 9− x2closest to the point (3, 9). See the figure3.4.
ANSWER. Using the usual distance formula, we find that the distance between the point (3, 9) and any point (x, y) is
d =
√
(x− 3)2+ (y− 9)2.
If the point (x, y) is on the parabola, it should satisfy the equation y = 9− x2. So the distance becomes d =
√
(x− 3)2+ (9− x2− 9)2 =
√
(x− 3)2+ x4
We want to find x minimizing the distance d. For easy computation, we find x minimizing d2. (Note that x = a minimizes d if and only if x = a minimizes d2.)
By the First Derivative Test and the techniques/concepts in previous sections, d2has the minimum value 5 at x = 1. Therefore, the closest point on the parabola is (x, y) = (1, 8) and the closest distance is√
5 . □
Figure 3.4: Closest Point
Example 3.7.8. Find the point on the given curve closest to the given point:
y = x2to the point (0, 1).
y = 4x + 7 to the origin.
Example 3.7.9. Find the points on the ellipse 4x2+ y2= 4 that are farthest away from the point (1, 0).
Remark 3.7.10(Important!). Check the function values at the critical numbers and at the endpoints. Do not simply assume (even by virtue of having only one critical number) that a given critical number corresponds to the extremum you are seeking.
□ 3.7.5 Soda Can & Highway.
Figure 3.5: Soda Can and Highway
Example 3.7.11. A soda can is to hold 12 fluid ounces. Find the dimensions that will minimize the amount of material used in its construction, assuming that the thickness of the material is uniform (i.e., the thickness of the aluminum is the same everywhere in the can). See the figure3.5.
ANSWER. Please read the textbook solution. □
Example 3.7.12. The state wants to build a new stretch of highway to link an existing bridge with a turnpike interchange, located 8 miles to the east and 8 miles to the south of the bridge. There is a 5–mile–wide stretch of marshland adjacent to the bridge that must be crossed. Given that the highway costs $10 million per mile to build over the marsh and only $7 million to build over dry land, how far to the east of the bridge should the highway be when it crosses out of the marsh? See the figure3.5.
ANSWER. Total cost is obtained as follows:
Total cost = 10× (distance across marsh) + 7 × (distance across dry land).
Letting x represent the distance in question, by the Pythagorean theorem, we get the cost function C(x) = 10√
x2+ 52 + 7√
(8− x)2+ 32. So we want to minimize this cost function. By the argument in previous sections and using a calculator, we obtain the minimum value C(xc)≈ $98.9 million at the critical number
xc≈ 3.560052. □
The problems in this section are applications of derivatives, especially the arguments used in curve sketching.
Definitely the problems are not easy to solve. As the first step of solving a word problem, you should be able to set up the mathematical expressions and for this work, you need to practice by solving lots of problems.
Solve Solve Solve Lots of Problems and Practice Practice Practice.