A clothing store sells 40 pairs of jeans daily at $30 each. The owner figures that for each $3 increase in price, 2 fewer pairs will be sold each day. What price should be charged to maximize profit?
To find the price that maximizes the profit, we need to first find the profit function. Let x be the number of $3 increases in price, then the price per pair of jeans is 30 + 3x. The number of pairs sold per day is 40 - 2x. The profit function is the product of the price and the quantity. Typically, we would have Profit = Revenue - Cost, where Cost would be something like $10 a pair; that is, 10(40 - 2x) if we considered it. However, since we don't know what Cost is and we don't need to know in order to solve this problem (since we need to maximize based on price), we will proceed without considering cost for this example:
P(x) = (30 + 3x)(40 − 2x)
P(x) = (1200 + 60x - 6x2)
P(x) = (-6x2 + 60x + 1200)
Here are 3 ways to solve it:
To find the maximum value of P(x), we need to find its vertex. The vertex of a quadratic function is given by:
x =-b
2a
where a and b are the coefficients of the quadratic and linear terms, respectively. In this case, a = -6 and b = 60, so:
x = -60
2(-6)
x = 5
Or you could complete the square to solve it:
P(x) = -6(x2 - 10x) + 1200
To complete the square, you take half of b, that is -10 and square it. So (-5)2 = 25. And when you add 25 multiplied by the -6 coefficent, you also have to add what will make it zero, which is 6(25):
P(x) = -6(x2 - 10x + 25) + 1200 + 6(25)
P(x) = -6(x - 5)2 + 1350
So, to get the initial part to 0, x would need to be 5.
This means that the price should be increased by $3 five times, or by $15 in total. The optimal price per pair of jeans is:
30 + 3(5) = 45
Or, if you are comfortable with Calculus, you could do a derivative:
P(x) = -6(x2 - 10x) + 1200
Get things in x terms with exponents:
P(x) = -6(x2 - 10x1) + 1200x0
Now to do the derivative, pay attention to what occurs with the exponents (where the become coefficents and then the exponent is reduced):
P'(x) = -6(2x1 - 1(10x0)) + 0(1200-1)
So P'(x) = -6(2x - 10)
For maximization, set it to zero and solve:
0 = -6(2x - 10)
0 = 2x - 10
10 = 2x
x = 5
Again 30 + 3(5) = 45
So in all thre ways we solved it, the value of x is 5. The price changes 5 times at $3 per change. The initial price was $30. If the price changes 5 times, then it is 5($3), which is $15. So, add that to the initial price of $30.
Therefore, the store should charge $45 per pair of jeans to maximize its profit.