# Word Problem#3

• May 12th 2006, 01:28 PM
Chrissy
Word Problem#3
A successful bake shop buys a very rich whipped-cream-covered cake from a supplier at $6.00 per cake and then sells the cakes for$10.00 each. The bake shop sells 200 cakes per week. Research shows that they can increase their profit if they increase the selling price. However, for every $0.50 increase, the sales will drop by five. Determine the selling price that will maximize profit. • May 13th 2006, 11:05 PM earboth Quote: Originally Posted by Chrissy A successful bake shop buys a very rich whipped-cream-covered cake from a supplier at$6.00 per cake and then sells the cakes for $10.00 each. The bake shop sells 200 cakes per week. Research shows that they can increase their profit if they increase the selling price. However, for every$0.50 increase, the sales will drop by five. Determine the selling price that will maximize profit.

Hello,

let x be the times which the price is increased. Then you get:

new price: 10 + .5*x
number of sold cakes: 200 - 5*x
money to be paid for the cakes: (200-5*x)*6
profit with respect to x : p(x)

Now you can form an equation

p(x) = (200-5*x)*(10+.5*x)-(200-5*x)*6

$p(x)=-2.5x^2+80x+800$

1. The graph of this function is a parabola where the vertex is at the maximum value. If you tranform this equation into the vertex form (?) of the parabola you'll get:
$p(x)=-2.5(x-16)^2+1440$. That means: If the price is increased 16 times by 0.5 $then the profit will be 1440$ instead of the initial 800 $. 2. You'll get the maximum of p if the 1rst derivative equals zero: ${dp(x) \over dx}=-5x+80$ . Let the RHS be zero, then x = 16. Plug in this value for x into p(x) and you'll get 1440$.

Bon appetit!

EB