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QuestionFind the profit function P(x)=−0.001x2+2.45x−510P(x)=-0.001 x^{2}+2.45 x-510 and determine the xx for max profit. What is the max profit in \$?

Studdy Solution

STEP 1

Assumptions1. The revenue function is given by R(x)=−0.001x+3xR(x)=-0.001 x^{}+3 x . The profit function is given by (x)=−0.001x+.45x−510.000(x)=-0.001 x^{}+.45 x-510.000
3. The profit function is the difference between the revenue and cost functions4. We are looking for the number of roast beef sandwiches that maximizes the profit

STEP 2

First, we need to find the derivative of the profit function to find the critical points. The critical points occur where the derivative is zero or undefined.
′(x)=ddx(−0.001x2+2.45x−510.000)'(x) = \frac{d}{dx}(-0.001 x^{2}+2.45 x-510.000)

STEP 3

Now, differentiate the profit function with respect to xx.
′(x)=−0.002x+2.45'(x) = -0.002x +2.45

STEP 4

Set the derivative equal to zero to find the critical points.
−0.002x+2.45=0-0.002x +2.45 =0

STEP 5

olve the equation for xx.
x=2.450.002x = \frac{2.45}{0.002}

STEP 6

Calculate the value of xx.
x=2.450.002=1225x = \frac{2.45}{0.002} =1225

STEP 7

Now that we have the number of sandwiches that gives a critical point, we need to determine whether this number maximizes or minimizes the profit. We can do this by taking the second derivative of the profit function.
′′(x)=d2dx2(−0.001x2+2.45x−510.000)''(x) = \frac{d^2}{dx^2}(-0.001 x^{2}+2.45 x-510.000)

STEP 8

Now, differentiate the derivative of the profit function with respect to xx.
′′(x)=−0.002''(x) = -0.002

STEP 9

Since the second derivative is negative, the function has a maximum at x=1225x =1225. This means that the store should make and sell1225 roast beef sandwiches each week to maximize profit.

STEP 10

Now, we need to find the maximum weekly profit. We can do this by plugging the value of xx that maximizes profit into the profit function.
(1225)=−0.001(1225)2+2.45(1225)−510.000(1225) = -0.001 (1225)^{2}+2.45 (1225)-510.000

STEP 11

Calculate the maximum weekly profit.
(1225)=−0.001(1225)+.45(1225)−510.000=$1225.00(1225) = -0.001 (1225)^{}+.45 (1225)-510.000 = \$1225.00The maximum weekly profit is $1225.00 when1225 sandwiches are sold.

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