Maximize Profit & Profit Calc for Jacket Prod.

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    Calculus
In summary, the conversation discusses finding the maximum profit for a manufacturer producing jackets. The revenue function is given as R(p)=4000[e^.01(p-100) + 1], with a cost of $50 per jacket. The production must be at least 100 and can't exceed 250. To find the maximum profit, one must find the value of p that maximizes the revenue function, which can be done by setting the derivative equal to zero. The profit is found by subtracting the cost from the revenue. The maximum profit is approximately $9500, achieved by producing and selling 250 jackets.
  • #1
kenny
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Pls Help! Calculus!

A manufacturer can produce jackets at a cost of $50 per jacket. If he produces p jackets weekly and all the jackets are sold, the revenue is R(p)=4000[e^.01(p-100) + 1]. Weekly production must be at least 100 and can't exceed 250... at wat price should the manufacturer sell the jackets to maximize profit... .and wat is the maximum weekly profit...
PLEASE HELP!
 
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  • #2
the e^(.01(p-100)) ... the .01(p-100) is to the base e
 
  • #3


Solve [tex] \frac {dR} {dp} = 0 [/tex] for [tex]p[/tex]

Then you can use the second derivative test to see if [tex]R(p)[/tex] is a maximum or minimum.

Doug
 
  • #4
read question carefully i need to find the max price of the jackets!... i don't understand this question
 
  • #5
The manufacturer wants to maximize profit thus he wants to maximize revenue.

What you need to do is find the number of jackets [tex]p[/tex] that maximizes revenue. The first and second derivatives of [tex]R(p)[/tex] help you find that value. Once you know the number of jackets that maximizes revenue then you can find the revenue that those jackets will generate. Then you can find the revenue that one jacket will generate which is the price. Once you know the price then you know the profit for each jacket.

First you must find the derivative:

[tex]\frac{dR}{dp}=\frac{d}{dp}4000(e^{.01(p-100)}+1)[/tex]

Then solve:

[tex]\frac{dR}{dp}=0[/tex]

for [tex]p[/tex]. Substituting the [tex]p[/tex] thus found into the original [tex]R(p)[/tex] will give you the revenue.

Then the price is:

[tex]\frac{R(p)}{p}[/tex]

and the profit is:

[tex]R(p)-50p[/tex]

Doug
 
  • #6
thanks doug... but i have a problem here... i found dR/dp= e^.01(p+100)... after i make that equal to 0... but e^x can't be zero!... so wat do i do?
 
  • #7
Then one of the endpoints of the interval takes the prize. Either

[tex]R(100)>R(250)[/tex]

or vice-versa.

Doug
 
  • #8
HOLD ON!

It's not the revenue you want to maximize but the profit!

The revenue is given as [tex]4000(e^{.01(p-100)}+1)[/tex] (which is obviously increasing for all x) and the cost function ("cost of $50 per jacket") 50p. The profit is [tex]4000(e^{.01(p-100)}+1)- 50p[/tex].

In order to find the maximum profit, you need to find
[tex]\frac{dP}{dp}=\frac{d}{dp}4000(e^{.01(p-100)}+1)-50p[/tex] and set it equal to zero.
That derivative is 400e^{.01(p-100)}- 50= 0 so
400e^{.01(p-100)}= 50

(Economists would say "marginal revenue equal marginal cost"- the revenue from selling "one more" jacket is equal to the cost of producing it.)

Hmm. The "marginal revenue" is still an exponential with a positive exponent so increasing: the maximum profit is still produced by producing and selling the largest possible number: 250 where the profit is about $9500.
 
  • #9
Although we come up with the same answer your reasoning is more universal, shall we say
 

1. How do you calculate profit for jacket production?

The profit for jacket production is calculated by subtracting the total cost of production from the total revenue generated from sales.

2. What factors affect the profit for jacket production?

The profit for jacket production can be affected by factors such as the cost of materials, labor costs, production volume, and selling price.

3. How can I maximize profit for jacket production?

To maximize profit for jacket production, you can try to reduce production costs, increase production volume, and optimize selling price based on market demand.

4. How do you use the Maximize Profit tool for jacket production?

The Maximize Profit tool for jacket production uses mathematical algorithms to analyze production costs and revenue to determine the optimal selling price for maximum profit.

5. Is the Maximize Profit tool accurate?

The Maximize Profit tool uses advanced mathematical models, but its accuracy ultimately depends on the accuracy of the data inputted for production costs and market demand. Regular updates and adjustments to the tool can also improve its accuracy.

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