1. Calculus Concepts 2/e LaTorre, Kenelly, Fetta, Harris, and Carpenter
Chapter 10
Analyzing Multivariable Change: Optimization
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2. Copyright © by Houghton Mifflin Company, All rights reserved.
Chapter 10 Key Concepts
Multivariable Critical Points
Multivariable Optimization
Constrained Optimization
The Method of Least Squares
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3. Multivariable Critical Points
Relative minimum: an output value smaller than any of those near it
Relative maximum: an output value larger than any of those near it
Absolute minimum: an output value smaller than all of those in the region
Absolute maximum: an output value larger than all of those in the region
Saddle point: an output that is a maximum of one cross section and a minimum of another
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4. Multivariable Critical Points: Example
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5. Multivariable Critical Points: Example
Contour curves near a saddle point curve away from the point.
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6. Multivariable Critical Points: Example
The percentage of sugar converted to olestra when the temperature is 150°C is shown below as a function of processing time and the ratio of peanut oil to sugar.
The saddle point occurs when time is 11 hours and the ratio is 11:1
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7. Critical Points: Exercise 10.1 #5
Make a dot on the contour plot at the approximate location of all critical points. Label points as relative maximum, relative minimum, or saddle point.
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8. Multivariable Optimization
Determinant Test
Let f be a continuous multivariable function with two input variables x and y. Let (a,b) be a point at which the first partial derivatives of f are both 0. The determinant of the second partials matrix evaluated at the point (a,b) is
If D(a,b) > 0 and fxx < 0 at (a,b), f(a,b) is a maximum.
If D(a,b) > 0 and fxx > 0 at (a,b), f(a,b) is a minimum.
If D(a,b) < 0 at (a,b), f has a saddle point at (a,b).
If D(a,b) = 0, the test fails.
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9. Multivariable Optimization: Example
The volume index for cake batter is given by
V(L, t) = -3.1L L - 0.1t t
when L grams of leavening is used and the cake is baked at 177°C for t minutes. Find the maximum volume possible and the conditions needed to achieve that volume.
VL = -6.2L = 0  L  3.6 grams of leavening
Vt = -0.2t = 0  t  26.5 minutes baking time
V(3.6, 26.5) 
Note: V = 100 is the uncooked batter volume.
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10. Multivariable Optimization: Example
The cross sections of V at L = 3.6 and t = 26.5 are both concave down which suggests the critical point is a maximum.
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11. Multivariable Optimization: Example
We must verify the volume is a maximum.
At (3.6, 26.5, 110.7),
VLL = -6.2
Vtt = -0.2
VLt = 0
VtL= 0
D(3.6, 26.5) > 0 and VLL < 0 so (3.6, 26.5, 110.7) is a maximum.
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12. Optimization: Exercise 10.2 #9
A restaurant mixes ground beef that costs $b per pound with pork sausage that costs $p per pound to make a meat mixture used on its pizza. The quarterly revenue from its pizza sales is given by
R(b,p) = 14b - 3b2 - bp - 2p2 + 12p thousand dollars.
Find the prices at which the restaurant should try to purchase the beef and sausage in order to maximize quarterly revenue.
Rb = b - p = 0
Rp = -b - 4p + 12 = 0
Solving the system of equations gives b  $1.91 and p  $ The critical point is (1.91, 2.52, 28.52).
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13. Optimization: Exercise 10.2 #9
Rbb = -6
Rbp = -1
Rpb = -1
Rpp = -4
D(1.91, 2.52) > 0 and Rbb < 0 so (1.91, 2.52, 28.52), is a maximum.
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14. Constrained Optimization
The minimum (or maximum) output of f subject to g(x,y) = c can be determined by finding the smallest (or largest) value M for which the constraint curve g(x,y) = c and the contour curve f(x,y) = M touch and then determining the point (x0, y0) where these two curves meet.
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15. Constrained Optimization
The Lagrange Multiplier
To find the optimal solution subject to the constraint we must solve the system of equations:
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16. Constrained Optimization: Example
A matrix manufacturing process has the Cobb-Douglas production function of
f(L,K) = 48.1L0.6K0.4 mattresses
where L represents the number of worker hours (in thousands) and K represents the amount invested in capital (in thousands of dollars). A plant manager has $98,000 to be invested in capital or spent on labor and the current average wage of an employee is $8 per hour. Write the budget constraint equation and find the capital and labor expenditures that will result in the greatest number of mattresses being produced subject to the budget constraint.
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17. Constrained Optimization: Example
The budget constraint equation is
g(L, K) = 8L + K thousand dollars
where L and K are as previously described.
The following equations must be satisfied at the point at which maximum production occurs.
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18. Constrained Optimization: Example
Solving the system yields K = 39.2 and L = Therefore, labor costs are (7.35 thousand worker hours)($8 per hour) = $58,800. The capital investment is $39,200. The production level is f(7.35, 39.2) =  690 mattresses.
To verify that a maximum occurred, we consider nearby points. Observe f(7.3, 39.6) = and f(7.4, 38.8) = which are both less than the extreme value. So a maximum occurred.
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19. Constrained Optimization: Exercise 10.3 #5
Find the optimal point of f(r,p) = 2r2 + rp - p2 + p under the constraint g(r,p) = 2r + 3p = 1. Identify the output as a maximum or minimum using close points.
Solving the system of equation yields, p = and r = We must determine if this is a maximum or minimum.
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20. Constrained Optimization: Exercise 10.3 #5
f( , 0.375) =
We select values of r close to and use g(r,p) = 1 to find the associated p value.
f(-0.062, ) =
f(-0.063, ) =
Evaluating f at these points we observe that f has a minimum (subject to the constraint) at r = and p =
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21. Constrained Optimization: Exercise 10.3 #5
The contour plot confirms the result.
f( , 0.375) =
f(-0.062, ) =
f(-0.063, ) =
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22. Method of Least Squares
The method of least squares is the procedure used to find the best-fitting line based on the criterion that the sum of squared errors is as small as possible. The linear model obtained by this method is called the least-squares line.
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23. Method of Least Squares: Example
The number of investment choices for people who invest in 401(k) plans at their place of employment is given by the data. A scatter plot and linear model are also shown.
Year
401(k) Plans
1990
1991
1992
1993
3.5
4.0
4.2
4.8
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24. Method of Least Squares: Example
The deviation values (d1, d2, d3, d4) measure the error between the model and the data points. We need to minimize the sum of the squared deviations.
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25. Method of Least Squares: Example
Finding fa and fb, setting them equal to zero, and solving the resultant system of equations yields a = 0.41 and b = Since faa(0.41, 3.51) > 0 and the determinant of the second partials matrix is 224 the error is a minimum.
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26. Method of Least Squares: Exercise 10.4 #3
Find the linear function that best fits the data. x y 10 20 3 2 1
fa = 1000a + 60b - 80 = 0
fb = 60a + 6b - 12 = 0
Solving the system yields, a = -0.1 and b = 3.
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27. Method of Least Squares: Exercise 10.4 #3y 10 20 3 2 1
So y = -0.1x + 3
Since f(-0.1, 3) = 0, the linear model is a perfect fit.
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