Asked by Ronny
To make pom poms in our school colors, we will have expenses of $100 to rent the Acme PomPom Plant, and then $0.25 per pom pom in materials. We believe that we can sell 500 pom poms if we charge $1.50. Assume the number sold is a linear function of the price. How much should we charge to maximize profit?
I have done this problem many times but I cannot find max profit. I keep fining a max income. I am lost and don't know what to do please help!
I have done this problem many times but I cannot find max profit. I keep fining a max income. I am lost and don't know what to do please help!
Answers
Answered by
MathMate
"Assume the number sold is a linear function of the price"
but no information is given on the price elasticity, so we will assume, in general, an increase in price of $1 will increase sales by m units.
Under normal supply-demand curve, m is necessarily negative, of the order of -200 or so.
With that in mind, and knowing that (500,1.50) is a point on the line sales versus price, we construct the sales(y)-price(x) relation as:
(y-500)=m(x-1.50)
therefore, at a price of x, we expect sales of
y=m(x-1.5)+500
Total revenue,
R=xy
Total profit
P=xy - cost
=xy - (0.25x+100)
=x(m(x-1.5)+500) - (0.25x+100)
To get the maximum profit, we differentiate profit with respect to price, and equate to zero to find the optimum price:
dp/dx = m*x+m*(x-1.5)+1999/4 =0
Solve for x:
x=(6*m-1999)/(8*m)
If the price elasticity m=-100,
x=2599/800=$3.25
if m=-200
x=3199/1600=$2
if m=-300
x=3799/2400=$1.58
...
etc.
but no information is given on the price elasticity, so we will assume, in general, an increase in price of $1 will increase sales by m units.
Under normal supply-demand curve, m is necessarily negative, of the order of -200 or so.
With that in mind, and knowing that (500,1.50) is a point on the line sales versus price, we construct the sales(y)-price(x) relation as:
(y-500)=m(x-1.50)
therefore, at a price of x, we expect sales of
y=m(x-1.5)+500
Total revenue,
R=xy
Total profit
P=xy - cost
=xy - (0.25x+100)
=x(m(x-1.5)+500) - (0.25x+100)
To get the maximum profit, we differentiate profit with respect to price, and equate to zero to find the optimum price:
dp/dx = m*x+m*(x-1.5)+1999/4 =0
Solve for x:
x=(6*m-1999)/(8*m)
If the price elasticity m=-100,
x=2599/800=$3.25
if m=-200
x=3199/1600=$2
if m=-300
x=3799/2400=$1.58
...
etc.
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