SOLUTION: a mathematically minded street trader with no overheads had found that the weekly volume of sales of a toy are approximately 100/p^2, where $p is the fixed price of the toy. The to

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: a mathematically minded street trader with no overheads had found that the weekly volume of sales of a toy are approximately 100/p^2, where $p is the fixed price of the toy. The to      Log On


   

Question 1175048: a mathematically minded street trader with no overheads had found that the weekly volume of sales of a toy are approximately 100/p^2, where $p is the fixed price of the toy. The toy costs the trader $0.15.
(a) find the level of p which maximizes profit
Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


The trader's cost is $0.15 for each toy.

The sale price of each toy is p.

The number of toys sold at a price of p is 100/p^2.

The total revenue is sale price times number of toys sold: p%28100%2Fp%5E2%29+=+100%2Fp

The total cost is the number of toys times the cost of each toy: 0.15%28100%2Fp%5E2%29+=+15%2Fp%5E2

Profit P (capital P, to distinguish from price p) is revenue minus cost: P%28p%29+=+100%2Fp+-+15%2Fp%5E2+=+100p%5E%28-1%29+-+15p%5E%28-2%29

The maximum profit is when the derivative of the profit function is 0.

dP%2Fdp+=+-100p%5E%28-2%29%2B30p%5E%28-3%29

0+=+-100%2Fp%5E2+%2B+30%2Fp%5E3
0+=+-100p%2B30
100p+=+30
p+=+30%2F100+=+0.30

The price p that maximizes profit is 0.30, or $0.30, or 30 cents.