Question

Consider the production function Y=X_1^(2/5) X_2^(1/5),P_(x_1 )=$6,P_(x_2 )=$3,P_y=$15. At which levels of X_1,X_2 and Y will profit be maximized?

Answer 1

The answer is X1 = 1, X2 = 1 and Y = 1

Profit = revenue - cost

π = price x quantity - X1PX1 - X2PX2

π = 15*(X1^2/5 X2^1/5) - 6X1 - 3X2

Profit is maximized for values of X1 and X2 at which the partial derivatives are 0

π'(X1) = 0

15*(2/5)X1^(-3/5)X2^(1/5) - 6 = 0

X1^(-3/5)X2^(1/5) = 1

π'(X2) = 0

15*(1/5)X1^(2/5)X2^(-1/5) - 3 = 0

X1^(2/5)X2^(-1/5) = 1

Divide them to get X2/X1 = 1 or X1 = X2

Production function is Y = X1^(2/5)X1^(1/5) = X1^(3/5)

Then we have X1^(-3/5)X2^(1/5) = 1 or X1^(-3/5)X1^(1/5) = 1. This gives X1 = 1 and X2 = 1

Finally, Y = 1^(3/5) = 1.

Hence X1 = X2 = Y1 = 1.