Question

A firm produces two types of earphones per year: x thousand of type A and y thousand of type B. The cost function (in thousands of dollars) is given by

C(x,y)=110+80x+130y
If the selling price of type A is p and the selling price of type B is q, the price-demand equations are
p=120-3x+y
q=210+x-7y
Determine how many of each type of earphone should be produced per year to maximize profit? What should be the selling prices p and q? What is the maximum profit?

Answer 1

• 9000 A and 7000 B should be produced
• p = 100, q = 170
• $350,000 is the maximum profit

Explanation:

We assume that "p" and "q" are given in dollars, so that the profit function (in thousands of dollars) can be written:

P = xp +yq -C(x, y)

Profit will be maximized when the partial derivatives of P with respect to x and y are both zero:

∂P/∂x = p +x(∂p/∂x) +y(∂q/∂x) -∂C/∂x = 0

(120 -3x +y) +x(-3) +y(1) -80 = 0

-6x +2y +40 = 0

3x -y = 20 . . . . put in standard form

and ...

∂P/∂y = x(∂p/∂y) +q +y(∂q/∂y) -∂C/∂y = 0

x(1) +(210 +x -7y) +y(-7) -(130) = 0

2x -14y +80 = 0

x -7y = -40 . . . . put in standard form

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The solution to these simultaneous equations can be found a variety of ways. Using Cramer's Rule, we have ...

x = ((-1)(-40) -(-7)(20))/(-1·1-(-7)(3)) = 180/20 = 9

y = (20(1) -(-40)(3))/20 = 140/20 = 7

9000 type A and 7000 type B earphones should be produced.

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The corresponding selling prices are ...

p = 120 -3(9) +7 = 100

q = 210 +9 -7(7) = 170

The selling prices should be ...

p = $100, q = $170

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The maximum profit is ...

P = xp +yq -C(x, y) = (9)(100) +(7)(170) -(110 +80(9) +130(7))

P = 9(100 -180) +7(170 -130) -110 = 180 +280 -110

P = 350

The maximum profit is $350,000.