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# a shoe retailer finds that she can sell 23 pairs per month when the price is \$43 per pair and that for each dollar by wich she lowers the price, her monthly sal

a shoe retailer finds that she can sell 23 pairs per month when the price is \$43 per pair and that for each dollar by which she lowers the price, her monthly sales increase by 5 pairs. If the shoes cost her \$30 a pair wholesale, what is the correct expression for her monthly profits in terms of her selling price per pair?

### 2 Answers by Expert Tutors

Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor
4.9 4.9 (409 lesson ratings) (409)
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The basic profit equation is:

P = (\$43)(23 pairs) - (\$30)(23 pairs) = (\$13)(23 pairs)

Lowering the price by \$x increases sales by 5x pairs of shoes, so the modified equation for monthly profit is:

P = (\$13-\$x)(23+5x) = 299 + 42x - 5x2

To find the maximum profit, take the derivative of P wrt x, set it zero, and solve for x:

dP/dx = 42 - 10x

0 = 42 - 10x

10x = 42

x = 4.2

The retailer will maximize her profit with a selling price of \$43 - \$4.20 = \$38.80.  She will sell 23 + 5(4.2) = 44 pairs of shoes.  Her profit will be \$387.20.

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You can also solve this algebraically by noting that the modified profit equation is an inverted parabola, so the maximum point will be the vertex, which is located at the point:

x = -b/2a = (-42)/(-10) = 4.2
PIYUSH L. | Maths tutoring for middle school to college maths studentsMaths tutoring for middle school to coll...
5.0 5.0 (6 lesson ratings) (6)
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Lets assume the profit will be maximized for the sale price of \$x

Profit per unit * Sales = Profit

Profit per unit = x-30

Now we know when the sale price is decreased by \$1 from 43, the sales of 23 increases by 5 additional units. So if we choose to have the sale price as \$x, then

sales = 23 + (43-x)*5

We can check the above equation by replacing x by 43, 42, 41 etc

Now using the above 3 equations we get,

(x-30)*(23+(43-x)*5)= Profit

(x-30)(238-5x)=profit

-5x2+150x+238x-5760 = profit

Now to maximize profit, we need to differentiate profit with respect to x and equal it to 0

-10x+388 = dP/dx = 0

x = 38.8

x is approximately 39

to make sure we have got the maximum, we will find d2P/d2x = -10
As the double derivative is negative, we are ensured that we have got the maximum

So to get the maximum profit, the sale price should be fixed at \$39

Max profit = (39-30)(238-5*39) = \$387