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P(x) = -0.4X^2 + fx - m

If $80 is charged for a design fee, and the monthly studio rent is 1,600; write an equation for the profit, P, in terms of x. Then how mch is the profit when 50 award designs are sold in a month? How many award designs must be sold in order to maximize the profit and show work algebraically and what is the maximun profit?

I need it explained in detail using the steps algebraically in all aspects.

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1 Answer

Given:     P(x) = -0.4x2 + fx - m

          design fee (f) = $80

          monthly studio rent (m) = $1,600

          # of award designs sold = x

     ==>     P(x) = -0.4x2 + 80x - 1600

Notice that this is a quadratic equation (standard form: P = ax2 + bx + c), which is parabolic. Parabolic functions open upwards when the leading coefficient ('a') is positive and open downwards when 'a' is negative. Since the leading coefficient in this equation (-0.4) is negative, the parabola opens downwards which means that the x-coordinate of the its vertex is the maximum value of the function and is given by:   x = -b/(2a). 

The profit when 50 award designs are sold in a month is determined as follows:

     x = 50

     P(50) = -0.4(50)2 + 80(50) - 1600

              = -0.4(2500) + 4000 - 1600

              = -1000 + 4000 - 1600

              = 1400

Therefore, the profit (P) when 50 award designs are sold is $1,400.

To maximize profit, the maximum value of x (award designs) must be sold. As mentioned above, the max value is given by the x-coordinte of the vertex of the parabola:     x = -b/(2a)   

    from the equation   ==>   b = 80 ,   a = -0.4

     x = -80/(2ยท-0.4) 

       = -80/-0.8 

       = 100

Thus, in order to maximize profit 100 award designs must be sold.

That maximum profit is given by P(x) when  x = 100:

     P(100) = -0.4(100)2 + 80(100) - 1600

               = -0.4(10000) + 8000 - 1600

               = -4000 + 8000 - 1600

               = 2,400

Thus, the maximum profit is $2,400.