# Max Profit

• Apr 5th 2009, 05:57 PM
sk8erboyla2004
Max Profit
This problem makes no sense to me

A store can sell 20 bikes per week at \$400. For each \$10 the store lowers the price by 2 more bicycles sold per week. each buke costs the store 200 dollars what price should be charged to maximize profit? how many bikes sold?

I have no clue how to formulate my cost and revenue functions from this
• Apr 5th 2009, 07:44 PM
Raldikuk
First you need to create the demand function to relate revenue to the price increase\decrease. To do this you will need to find another point to calculate the slope. I picked the point 22, 390 (\$10 decrease 2 bikes sold increase).

Now I calculate the slope (390-400)/(22-20)=-5. Now you can put this into slope intercept form with the original pts. y-400=-5(x-20) sim: y=-5x+300.

Now we can create the primary function. Profit is Revenue - Cost. What is the revenue in this case? It is number of units sold multiplied by price. This is x*y. Y is related to x via the demand function we just created. What is the cost? The cost is a given, and it is \$200. However, we must remember to multiply it by the number of units sold; so for the function the cost is 200x.

So the primary function will be P(x)=x*(-5x+300)-200x. sim: -5x^2+100x

Now we can differentiate and solve for zero to find critical values and possible mins and maxes.

-10x+100=0
-10x=-100
x=10

Maximum profit will be made when 10 units are sold. Given our earlier relationship we know that the cost per unit will be \$250. The cost will be \$2000, so the max profit that can be made is \$500.