5.1 - 5.4: Curve Sketching
5.5: Absolute Extrema
Recall that given a function f ( x ), the absolute (global) minimum is the smallest value of f on its domain, and the absolute (global) maximum is the largest value of f on its domain.
In plural, extrema are the maximum and minimum of f .
The Extreme Value Theorem: If f is continuous on a closed interval [ a, b ], then f attains its absolute maximum value f ( c ) and absolute minimum value f ( d ) at some number c and d in [ a, b ].
Note:
• An absolute extremum for a continuous function on a closed interval can be a relative extremum or one of the end points of the interval.
• An extreme value can be taken on more than once.
• A function may not have an absolute maximum or minimum if it is not continuous or the interval is not closed and bounded.
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Procedure for finding absolute extrema: Let f be a continuous function on a closed and bounded interval
[ a, b ].
1) Locate all critical values on ( a, b ).
2) Evaluate f ( x ) at those the critical values and also at a and b .
3) The absolute maximum of f ( x ) on [ a, b ] will be the largest number found in Step 2, while the absolute minimum of f ( x ) on [ a, b ] will be the smallest number found in Step 2.
Example 1.
Locate the value(s) where each function attains an absolute maximum and the value(s) where the function attains an absolute minimum, if they exist, of the given function on the given interval.
a) f ( x ) = − x
2
+ 2 x + 3 on [0 , 4] .
b) f ( x ) = x 3 + x 2 − 5 x + 1 on [ − 2 , 2] .
1 c) gx ) = x on ( −∞ , 5] .
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5.6: Optimization and Modeling
Six-Step Mathematical Modeling for Optimization:
1. Identify the variables (unknown quantities) and denote each with a symbol.
2. Identify the variable that is to be maximized or minimized and express it as a function of the other variables.
3. Find any relationships between the variables.
4. Express the quantity to be maximized or minimized as a function of only one variable. To do this, you might need to solve for one of the variables in one of the equations in Step 3 and substitute this into the formula found in Step 2.
5. Determine the interval over which the function is to be minimized or maximized.
6. Steps 1 through 5 leave you with a function of a single variable on a certain interval to be maximized or minimized. Proceed now as in previous sections.
Example 2.
Find two positive number whose product is 144 and whose sum is a minimum.
Example 3.
Find two number whose sum is 90 and whose product is maximum.
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Example 4.
A farmer with 1000 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible area of the four pens?
Example 5.
A box with a square base and open top must have a volume of 50,000 cm 3 . Find the dimensions of the box that minimize the amount of material used.
Example 6.
A box with an open top is to be constructed from a square piece of cardboard, 5 ft wide, by cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a box can have.
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Business Applications:
Recall 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 ( x ) with respect to x . I.e., the marginal cost function is the derivative C
0
( x ) of the cost function.
The average cost function is the cost per unit when x units produced.
Example 7.
Find the derivative of
¯
( x ) .
¯
( x ) =
C ( x ) x
Remark: From that example, the average cost is minimized at the value x at which ¯ ( x ) = C
0
( x ) .
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Example 8.
A company estimates that the cost (in dollars) of producing x items is
C ( x ) = 2 x
3
2
+ x
2
1000
.
a) Find the cost of producing 900 items.
b) Find the average cost of producing 900 items.
c) Find the marginal cost of producing 900 items.
d) At what production level will the average cost be lowest? What is this minimum average cost?
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Let p ( x ) be the price-demand function. Recall that if x units are sold and the price per unit is p ( x ), then the total revenue is R ( x ) = xp ( x ). The derivative R
0 of the revenue function is the marginal revenue function, i.e. the rate of change of revenue with respect to the number of units sold.
If x units are sold, then the total profit is P ( x ) = R ( x ) − C ( x ) and the marginal profit function P
0 is the derivative of the profit function. In order to maximize profit, we look for the critical value of function P by solving:
P
0
( x ) = R
0
( x ) − C
0
( x ) = p ( x ) + xp
0
( x ) − C
0
( x ) = 0 .
The profit is maximized at the solution of that equation.
Example 9.
Determine the production level that will maximize the profit for a company with cost and demand functions C ( x ) = 1200 + 25 x − 0 .
0001 x 2 and p ( x ) = 50 − 0 .
001 x.
Example 10.
Let the cost for a certain product be given by C ( x ) = 0 .
001 x 3 − 0 .
3 x 2 + 6 x + 1000 . Find the production level at which the marginal cost function starts to increase.
Example 11.
The manager of a 200-unit apartment complex knows from experience that all units will be occupied if the rent is $ 500 per month. A market survey suggests that, on the average, one additional unit will remain vacant for each $ 5 increase in rent. What rent should the manager charge to maximize revenue?
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