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FOR SECTION 33 1. (a) Solve the problem max 100- x2-y -z subject to x+2y+z = a. (b) Compute the optimal value function f*(a) and verify that (9) holds.

FOR SECTION 33 1. (a) Solve the problem max 100- x2-y -z subject to x+2y+z = a. (b) Compute the optimal value function f*(a) and verify that (9) holds.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Can you help with question 1? Item (9) is attached for reference. 

86 7 PROBLEMS EOR SECTION33 1. (a) Solve the problem max 100 – x² - y² – z² subject to x+2y + z = a. (b) Compute the optimal value function f*(a) and verify that (9) holds. SM 2. (a) Solve the problem max x + 4y + z subject to x+y +z = 216 and x+2y+ 3z = 0 (b) Change the first constraint to x² + y² + z² = 215 and the second to x + 2y + 3z = 0.1. Estimate the corresponding change in the maximum value by using (30). 3. (a) Solve the problem max e +y +z subject to I = 2 + k + x I =2+ z+ x (b) Replace the constraints by x + y + z = 1.02 and x + y² + z? = 0.98. What is the approximate change in optimal value of the objective function? SM 4. (a) Solve the utility maximizing problem (assuming m > 4) max U(x1, x2) = In(1+ x1) + In(1+ x2) subject to 2x1 + 3x2 = m (b) With U* (m) as the indirect utility function, show that dU*/dm = . je Mous 1257 AM 4/25/2021 BANG & OLUFSE prt sc delete home 144 pua wnu lock backspace 9. 8. 6 10 D. home· dn 60
Transcribed Image Text:86 7 PROBLEMS EOR SECTION33 1. (a) Solve the problem max 100 – x² - y² – z² subject to x+2y + z = a. (b) Compute the optimal value function f*(a) and verify that (9) holds. SM 2. (a) Solve the problem max x + 4y + z subject to x+y +z = 216 and x+2y+ 3z = 0 (b) Change the first constraint to x² + y² + z² = 215 and the second to x + 2y + 3z = 0.1. Estimate the corresponding change in the maximum value by using (30). 3. (a) Solve the problem max e +y +z subject to I = 2 + k + x I =2+ z+ x (b) Replace the constraints by x + y + z = 1.02 and x + y² + z? = 0.98. What is the approximate change in optimal value of the objective function? SM 4. (a) Solve the utility maximizing problem (assuming m > 4) max U(x1, x2) = In(1+ x1) + In(1+ x2) subject to 2x1 + 3x2 = m (b) With U* (m) as the indirect utility function, show that dU*/dm = . je Mous 1257 AM 4/25/2021 BANG & OLUFSE prt sc delete home 144 pua wnu lock backspace 9. 8. 6 10 D. home· dn 60
Interpreting the Lagrange Multipliers Equation (6) can be written as (q),fe = 2; (b), j3D1,..., m (6) qe This tells us that the Lagrange multiplier ^; = 1; (b) for the jth constraint is the rate at which the optimal value of the objective function changes w.r.t. changes in the constant bj. Suppose, for instance, that f* (b) is the maximum profit that a firm can obtain from a production process when b1, . .., bm are the available quantities of m different resources. Then af*(b)/ab; is the marginal profit that the firm can earn per extra unit of resource j, which is therefore the firm's marginal willingness to pay for this resource. Equațion (9) tells us that this marginal willingness to pay is equal to the Lagrange multiplier 2; for the corresponding resource constraint whose right-hand side in (1) is b¡. If the firm could buy more of this resource at a price below 2¡ per unit, it could earn more profit by doing so; but if the price exceeds Aj, the firm could increase its overall profit by selling a small enough quantity of the resource at this price because the revenue from the sale would exceed the profit from production that is sacrificed by giving up the resource. In economics, the number ); (b) is referred to a shadow price (or demand price) of the resource j. It is a shadow rather than an actual price because it need not correspond to a market price. Indeed, the resource may be something unique like a particular entrepreneur's time for which there is not even a market that could determine its actual price. Note that the Lagrange multipliers for problem (1) may well be negative, so that an increase in b; can lead to a decrease in the value function. ' If db1, ..., dbm are small in absolute value, then, according to the linear approximation formula, f*(b+db)– f*(b) × (af*(b)/ab¡) dbi+…+(af*(b)/abm) dbm and, using (9), (10) "ap (q) Y + + !qp (q) 'y (q),S – (qp + q),f Show all 12:58 AM 4/25/2021
Transcribed Image Text:Interpreting the Lagrange Multipliers Equation (6) can be written as (q),fe = 2; (b), j3D1,..., m (6) qe This tells us that the Lagrange multiplier ^; = 1; (b) for the jth constraint is the rate at which the optimal value of the objective function changes w.r.t. changes in the constant bj. Suppose, for instance, that f* (b) is the maximum profit that a firm can obtain from a production process when b1, . .., bm are the available quantities of m different resources. Then af*(b)/ab; is the marginal profit that the firm can earn per extra unit of resource j, which is therefore the firm's marginal willingness to pay for this resource. Equațion (9) tells us that this marginal willingness to pay is equal to the Lagrange multiplier 2; for the corresponding resource constraint whose right-hand side in (1) is b¡. If the firm could buy more of this resource at a price below 2¡ per unit, it could earn more profit by doing so; but if the price exceeds Aj, the firm could increase its overall profit by selling a small enough quantity of the resource at this price because the revenue from the sale would exceed the profit from production that is sacrificed by giving up the resource. In economics, the number ); (b) is referred to a shadow price (or demand price) of the resource j. It is a shadow rather than an actual price because it need not correspond to a market price. Indeed, the resource may be something unique like a particular entrepreneur's time for which there is not even a market that could determine its actual price. Note that the Lagrange multipliers for problem (1) may well be negative, so that an increase in b; can lead to a decrease in the value function. ' If db1, ..., dbm are small in absolute value, then, according to the linear approximation formula, f*(b+db)– f*(b) × (af*(b)/ab¡) dbi+…+(af*(b)/abm) dbm and, using (9), (10) "ap (q) Y + + !qp (q) 'y (q),S – (qp + q),f Show all 12:58 AM 4/25/2021
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