SOLUTION: How do you find the vertex when you have a word problem? The word problem: A company sells 80 team jackets a week for $30 each. The product manager estimates that for each $3

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: How do you find the vertex when you have a word problem? The word problem: A company sells 80 team jackets a week for $30 each. The product manager estimates that for each $3       Log On


   



Question 41123: How do you find the vertex when you have a word problem?
The word problem: A company sells 80 team jackets a week for $30 each. The product manager estimates that for each $3 increase in price, the company will sell 5 fewer jackets per week. What price should be charged per jacket to maximize the profit?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
If they're looking to maximize profit, the curve should
have a peak instead of a dip.
Let x = the number of $3 increases in price.
According to the problem, 5x fewer jackets per week will be
sold with x number of $3 increases in price
The price per jacket is
30+%2B+3x
The corresponding number of jackets sold is
80+-+5x
Let income from sales of jackets = I
%2830+%2B+3x%29%2880+-+5x%29+=+I
I assume if income is maximized, profit will also be maximized
factor out 3 from (30 - 3x) and 5 from (80 - 5x)
5%2A3%2A%2810+%2B+x%29%2816+-+x%29+=+I
160+%2B+16x+-10x+-+x%5E2+=+I%2F15
-x%5E2+%2B+6x+%2B+160+=+I%2F15
The -x%5E2 signals that the curve has a positive peak
use
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-6+%2B-+sqrt%28+%286%5E2%29-4%2A%28-1%29%2A160+%29%29%2F%282%2A%28-1%29%29+
x+=+%28-6+%2B-+sqrt%28+%286%5E2%29-4%2A%28-1%29%2A160+%29%29%2F%282%2A%28-1%29%29+
%28-6%2F-2%29+%2B-sqrt%2836+%2B+640%29%2F-2
3+%2B-sqrt%28676%29%2F-2
3+%2B+26%2F2, 3+-+26%2F2
3 + 13 and 3 - 13 are the answers, or
16 and -10
This is where the meaaning of vertex comes in. 16 and -10
are the solutions to -x%5E2+%2B+6x+%2B+160+=+0 which gives
us the ROOTS of the equation, the values of x that make it zero,
or in this case, NO income.
So, we don't want the roots, what do we want?
We want the value of x midway between the roots
3 is midway between -10 and 16.
-10 is at 3 -13
16 is at 3 + 13
What price should be charged per jacket to maximize the profit?
That's what the problem wants
%2830+%2B+3x%29 is the price per jacket
30+%2B+3%2A3+=+39
$39 per jacket is the answer
Check the answer. What's the Income if $39/jacket is charged?
%2830+%2B+3x%29%2880+-+5x%29+=+I%2F15
%2830+%2B+3%2A3%29%2880+-+5%2A3%29+=+I%2F15
%2839%29%2865%29+=+I%2F15
2535+=+I%2F15
Reduce x slightly (even though you can't sell a fraction of a jacket)
to 2.9
%2830+%2B+3%2A2.9%29%2880+-+5%2A2.9%29+=+I%2F15
%2830+%2B+8.7%29%2880+-+14.5%29+=+I%2F15
38.7%29%2865.5%29+=+I%2F15
2534.85+=+I%2F15
The income dropped slightly with a .1 decrease in x
Increase x slightly to 3.1
%2830+%2B+3%2A3.1%29%2880+-+5%2A3.1%29+=+I%2F15
%2830+%2B+9.3%29%2880+-+15.5%29+=+I%2F15
%2839.3%29%2864.5%29+=+I%2F15
2534.85+=+I%2F15
The income dropped slightly with a .1 increase in x, as expected,
so the peak income really is at x = 3
and $39 per jacket will maximize the weekly profit