Question

4. For the following total profit function of a firm, where X and Y are two goods sold by the firm:

Profit = 226X – 5X2 –XY -2.5Y2 +150Y -210

(a) Determine the levels of output of both goods at which the firm maximizes total profit.

(b) Calculate the profit.

c) Do problem 4 again, but this time with a constraint of X + Y =40. a. To work problem 5 use the constraint equation to find an equation for X in terms of Y (X = 40 – Y) or an equation for Y in terms of X. b. Substitute the equation into the total profit equation at every place you find X (or Y) c. Set the derivative equal to zero and solve for X (or Y) d. Plug the result from part c into the constraint equation to get the other variable.

(please show work for each step)

Answer 1

The total profit function of a firm is Profit = 226X – 5X2 –XY -2.5Y2 +150Y -210

(a) Determine the levels of output of both goods at which the firm maximizes total profit.

Find the first order partial derivatives of profit function with respect to X and Y

Π’(X) = 0

226 – 10X – Y = 0…….(1)

– X – 5Y + 150 = 0………..(2)

Solve the two equations using the value of Y from first equation (1) as Y = 226 – 10X and place it in (2)

– X – 5(226 – 10X) + 150 = 0

– X – 1130 + 50X + 150 = 0

X* = 20 and Y* = 26

(b) Calculate the profit.

The profit is Π = 226*20 – 5*20^2 –20*26 – 2.5*26^2 + 150*26 – 210 = 4000 Π =

c) Do problem 4 again, but this time with a constraint of X + Y =40.

Now that X is equal to 40 – Y, profit function becomes

Π =226(40 – Y) – 5(40 – Y)^2 –(40 - Y)Y – 2.5Y^2 +150Y – 210

= 9040 – 226Y – 8000 – 5Y^2 + 400Y – 40Y + Y^2 – 2.5Y^2 + 150Y – 210

= 830 – 284Y – 6.5Y^2

Profit is maximized when Π’(X) = 0

284 – 13Y = 0

Y = 21.85 and X = 40 – 21.85 = 18.15

Profit is 3932.154