SOLUTION: Jack & Jill charge $20 for a custom painted water pail. They currently sell 80 water pails a month. They estimate that for each $1 decrease in the cost of the pails, they could sel

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Jack & Jill charge $20 for a custom painted water pail. They currently sell 80 water pails a month. They estimate that for each $1 decrease in the cost of the pails, they could sel      Log On


   

Question 1060581: Jack & Jill charge $20 for a custom painted water pail. They currently sell 80 water pails a month. They estimate that for each $1 decrease in the cost of the pails, they could sell 15 more pails a month. What price will maximize Jack & Jill's income? If they charges this price, how much income should they expect?
Found 3 solutions by josmiceli, josgarithmetic, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +n+ = the number of $1 decreases in price
Let +I+ = the total income from sales of water pails
-------------------------------
[ income ] = [ price per pail ] x [ number of pails sold ]
+I+=+%28+20+-+1%2An+%29%2A%28+80+%2B+15n+%29+
+I+=+1600+-+80n+%2B+300n+-+15n%5E2+
+I+=+-15n%5E2+%2B+220n+%2B+1600+
The n-value of the vertex ( max in this case ) is:
+n%5Bv%5D+=+-b%2F%282a%29+
+a+=+-15+
+b+=+220+
+n%5Bv%5D+=+-220%2F%28+2%2A%28-15%29+%29+
+n%5Bv%5D+=+7.333+
-----------------------------
Going back to original equation:
[ price/pail ] = +20+-+n+
The price that maximizes income is:
+20+-+7.333+=+12.333+
$12.33
-----------------------------
The max income at this price is:
+I+=+-15n%5E2+%2B+220n+%2B+1600+
+I%5Bmax%5D+=+-15%2A7.333%5E2+%2B+220%2A7.333+%2B+1600+
+I%5Bmax%5D+=+-806.593+%2B+1613.26+%2B+1600+
+I%5Bmax%5D+=+2406.67+
$2,406.67
-----------------------
here's the plot:
+graph%28+600%2C+400%2C+-2%2C+25%2C+-300%2C+2600%2C+-15x%5E2+%2B+220x+%2B+1600+%29+



Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
r, income
x, how many pails sold
p, price per pail
r=px

n, each dollar decrease in price
PRICE    x
 20      80
 20-1    80+15
 20-2    80+2*15
 20-3    80+3*15
 20-n    80+n*15 or 80+15n

highlight%28r=%2820-n%29%2880%2B15n%29%29

r is a parabolic function with zeros at %2820-n%29%2880%2B15n%29=0, and the MAXIMUM revenue will be exactly in the middle of the two zeros for n.

--
20-n=0
n=20
-
80%2B15n=0
15n=-80
n=-80%2F15
n=-5%261%2F3
-
MIDDLE n VALUE
%2820-5%261%2F3%29%2F2
highlight%28n=7%261%2F3%29----------how many dollars LESS THAN 20, for maximum revenue.

If Jack and Jill want a price to the nearest penny, then this would be $7.33 per pail.
Revenue would be r=7.33%2880%2B7.33%2A15%29=highlight%281392.33%29.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Jack & Jill charge $20 for a custom painted water pail. They currently sell 80 water pails a month. They estimate that for each $1 decrease in the cost of the pails, they could sell 15 more pails a month. What price will maximize Jack & Jill's income? If they charges this price, how much income should they expect?
matrix%281%2C4%2C+7%261%2F3%2C+%22%241%22%2C+price%2C+reductions%29 for a reduction of $7.33, or a reduced price of highlight_green%28%22%2412.67%22%29 will maximize their income.

At a reduced price of $12.67, MAXIMUM income is: highlight_green%28%22%242%2C406.67%22%29

Let no-one tell you otherwise!!