Suppose that an individual consumes three goods -- food, clothing, automobiles.

Question

Suppose that an individual consumes three goods -- food, clothing, automobiles. Denote the quantities of these goods consumed by X, Y, and Z respectively. Suppose the individual utility function is given by U = 5 ln x + 4 ln y = ln (1+z) and the prices of the goods are given by Price X = $1 . Price Y = $2. and Price Z = $2,000.
Let the individual's total income be $56,000. Automobiles, however, have the property that they must be bought in discrete untis (that is, Z must equal 0, 1, 2, and so on). It is impossible in this simple model to buy one half a car. Given these contraints, how will an individual choose to allocate income so as to maximize utility?

Explanation / Answer

U = 5 ln x + 4 ln y = ln (1+z) I think you mean "U = 5 ln x + 4 ln y + ln (1+z)" This appears to be a typo. Please let me know in the comments section if it is not. If this isn't a typo, this becomes a very strange and difficult problem. max 5 ln x + 4 ln y + ln (1+z) s.t. x + 2y + 2000z = 56000 The Lagrange function is: L = 5ln x + 4ln y + ln(1+z) - v(x + 2y + 2000z - 56000), where v is a Lagrange multiplier. Take first order conditions. dL/dx: 5/x = v (1) dL/dy: 4/y = 2v (2) dL/dz: 1/(1+z) = 2000v (3) dL/dv: x + 2y + 2000z = 56000 (4) First solve for z by putting everything in terms of z Divide (1) by (3) 5(1+z)/x = 1/2000 x = 10000(1+z) Divide (2) by (3) 4(1+z)/y = 1/1000 y = 4000(1+z) Substitute these into (4) x + 2y + 2000z = 56000 10000(1+z) + 2*4000(1+z) + 2000z = 56000 10000 + 10000z + 8000 + 8000z + 2000z = 56000 z(10000 + 8000 + 2000) = 56000 - 10000 - 8000 z = (56000 - 10000 - 8000)/(10000 + 8000 + 2000) z = 1.9 But you can't have 1.9 cars, remember! You can only have 1 or 2 cars. So, what we have to do now is see if 1 or 2 cars will give us the higher utility. Scenario 1: We consume 1 car. The new Lagrange function is: L = 5ln x + 4ln y + ln(2) - v(x + 2y + 2000 - 56000) Take first order conditions. dL/dx: 5/x = v (1) dL/dy: 4/y = 2v (2) dL/dv: x + 2y + 2000 = 56000 (3) Divide (1) by (2) 5y/4x = 1/2 x = (10/4)y y = (4/10)x Use (3) to solve for x and y. x + 2y + 2000 = 56000 (10/4)y + 2y + 2000 = 56000 y(10/4 + 2) + 2000 = 56000 y = (56000 - 2000)/(10/4 + 2) y = 12000 x + 2y + 2000 = 56000 x + 2*12000 + 2000 = 56000 x = 56000 - 2*12000 - 2000 x = 30000 And the utility of this bundle is: U1 = 5ln x + 4ln y + ln(2) U1 = 5*ln(30000) + 4*ln(12000) + ln(2) U1 = 89.8085582 Scenario 2: We consume 2 cars. The new Lagrange function is: L = 5ln x + 4ln y + ln(3) - v(x + 2y + 2000*2 - 56000) Take first order conditions. dL/dx: 5/x = v (1) dL/dy: 4/y = 2v (2) dL/dv: x + 2y + 2000*2 = 56000 (3) Divide (1) by (2) 5y/4x = 1/2 x = (10/4)y y = (4/10)x Use (3) to solve for x and y. x + 2y + 2000*2 = 56000 (10/4)y + 2y + 2000*2 = 56000 y(10/4 + 2) + 2000*2 = 56000 y = (56000 - 2000*2)/(10/4 + 2) y = 11555.56 x + 2y + 2000*2 = 56000 x + 2*11555.56 + 2000*2 = 56000 x = 56000 - 2*11555.56 - 2000*2 x = 28 888.88 And the utility of this bundle is: U2 = 5ln x + 4ln y + ln(3) U2 = 5*ln(28888.88) + 4*ln(11555.56) + ln(3) U2 = 89.8743604 U1 = 89.8085582 U2 = 89.8743604 U2 > U1 So, the optimal bundle is obtained under scenario 2! (x* , y* , z*) = (28888.88 , 11555.56 , 2) We would be best off if we purchased 1.9 cars, but we can't. So we go ahead and purchase 2 cars and reduce our consumption of food and clothing by just enough so that we can afford everything.

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