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?

Step-by-step 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.

(-1,-0.5)

step-by-step explanation: