3.7 Optimization Problems
Optimization Problems
In solving such practical problems the greatest challenge is often to convert the word problem into a mathematical
optimization problem by setting up the function that is to be maximized or minimized.
Let’s recall the problem-solving principles.
Optimization Problems
Steps In Solving Optimization Problems
1. Understand the Problem The first step is to read the problem carefully until it is clearly understood. Ask
yourself: What is the unknown? What are the given quantities? What are the given conditions?
2. Draw a Diagram In most problems it is useful to draw a diagram and identify the given and required quantities on the diagram.
3. Introduce Notation Assign a symbol to the quantity that
Optimization Problems
Also select symbols (a, b, c, . . . , x, y) for other unknown quantities and label the diagram with these symbols. It may help to use initials as suggestive symbols—for example, A for area, h for height, t for time.
4. Express Q in terms of some of the other symbols from Step 3.
5. If Q has been expressed as a function of more than one variable in Step 4, use the given information to find
relationships (in the form of equations) among these variables. Then use these equations to eliminate all but one of the variables in the expression for Q.
Optimization Problems
Thus Q will be expressed as a function of one variable x, say, Q = f(x). Write the domain of this function in the
given context.
6. Use the previous methods to find the absolute maximum or minimum value of f. In particular, if the domain of f is a closed interval, then the Closed Interval Method can be used.
Example 1
A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?
Example 1 – Solution
In order to get a feeling for what is happening in this problem, let’s experiment with some specific cases.
Figure 1 (not to scale) shows three possible ways of laying out the 2400 ft of fencing.
Figure 1
Example 1 – Solution
We see that when we try shallow, wide fields or deep, narrow fields, we get relatively small areas. It seems
plausible that there is some intermediate configuration that produces the largest area.
Figure 2 illustrates the general case. We wish to maximize the area A of the rectangle.
cont’d
Figure 2
Example 1 – Solution
Let x and y be the depth and width of the rectangle (in feet).
Then we express A in terms of x and y:
A = xy
We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the given information that the total length of the fencing is 2400 ft.
Thus
2x + y = 2400
From this equation we have y = 2400 – 2x, which gives
cont’d
Example 1 – Solution
Note that the largest x can be is 1200 (this uses all the fence for the depth and none for the width) and x can’t be negative, so the function that we wish to maximize is
A(x) = 2400x – 2x2 0 ≤ x ≤ 1200
The derivative is A′(x) = 2400 – 4x, so to find the critical numbers we solve the equation
2400 – 4x = 0 which gives x = 600.
The maximum value of A must occur either at this critical number or at an endpoint of the interval.
cont’d
Example 1 – Solution
Since A(0) = 0, A(600) = 720,000, and A(1200) = 0, the Closed Interval Method gives the maximum value as
A(600) = 720,000.
[Alternatively, we could have observed that A′′(x) = –4 < 0 for all x, so A is always concave downward and the local maximum at x = 600 must be an absolute maximum.]
The corresponding y-value is y = 2400 – 2(600) = 1200; so the rectangular field should be 600 ft deep and 1200 ft wide.
cont’d
Optimization Problems
Applications to Business and
Economics
Applications to Business and Economics
We know that if C(x), the cost function, is the cost of producing x units of a certain product, then the marginal cost is the rate of change of C with respect to x.
In other words, the marginal cost function is the derivative, C′(x), of the cost function.
Now let’s consider marketing. Let p(x) be the price per unit that the company can charge if it sells x units.
Then p is called the demand function (or price function) and we would expect it to be a decreasing function of x.
Applications to Business and Economics
If x units are sold and the price per unit is p(x), then the total revenue is
R(x) = quantity × price = xp(x) and R is called the revenue function.
The derivative R′ of the revenue function is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold.
Applications to Business and Economics
If x units are sold, then the total profit is P(x) = R(x) – C(x)
and P is called the profit function.
The marginal profit function is P′, the derivative of the profit function.
Example 6
A store has been selling 200 flat-screen TVs a week at
$350 each. A market survey indicates that for each $10 rebate offered to buyers, the number of TVs sold will
increase by 20 a week. Find the demand function and the revenue function. How large a rebate should the store offer to maximize its revenue?
Solution:
If x is the number of TVs sold per week, then the weekly increase in sales is x – 200.
For each increase of 20 units sold, the price is decreased
Example 6 – Solution
So for each additional unit sold, the decrease in price will be and the demand function is
p(x) = 350 – (x – 200) = 450 – x The revenue function is
R(x) = xp (x) = 450x – x2
Since R′(x) = 450 – x, we see that R′(x) = 0 when x = 450.
This value of x gives an absolute maximum by the First
Derivative Test (or simply by observing that the graph of R is a parabola that opens downward).
cont’d
Example 6 – Solution
The corresponding price is
p(450) = 450 – (450) = 225
and the rebate is 350 – 225 = 125.
Therefore, to maximize revenue, the store should offer a rebate of $125.
cont’d
