max value question

Question

Caresia is an artist who works at a her shop drawing “watercolor portraits.” She charges $20 per portrait and she has been averaging 30 portraits per week. She decides to increase the price, but realizes that for every one dollar increase she will lose one sale per week. If materials cost her $10 per portrait, what should she set the price at in order to maximize her profit?

Answer 1

Let x = number of $1 increases in price

 

Profit = Revenue - Cost

 

Revenue if price per painting is increased $x

 

   = (price per painting)(number sold)

 

   = (20 + x)(30 - x)

 

Cost = (cost per painting)(number sold)

 

       = 10(30 - x)

 

Profit = P(x) = (20+x)(30-x) - 10(30-x)

 

                  = -x² + 10x + 600 - 300 + 10x

 

                  = -x² + 20x + 300

 

The graph of the profit function is an inverted parabola with maximum when x = -20/((2)(-1)) = 10.

 

So, the price per painting should be increased by $10 in order to maximize profit.  

Answer 2

Denote the new price by P

 

Denote the new quantity by Q

 

Denote the cost by C

 

Then:

 

P=20+x

 

Q=30-x

 

C=10(30-x)

 

Profit is PQ-C

 

Profit = -x^2+10x+300

 

d/dx Profit = -2x+10=0

 

x=5

 

New price should be 25.

 

Note that profit function is an inverted parabola, so x=5 is a maximum.