QuestionThe profit function is $P(x)=-0.001 x^{2}+2.4 x-530$ . Find $P(x)$ in simplified form and determine $x$ for max profit.

Studdy Solution

STEP 1

Assumptions1. The revenue function is given by $R(x)=-0.001 x^{}+3 x$ . The profit function is given by $(x)=-0.001 x^{}+.4 x-530.00$
3. The profit function is the difference between the revenue function and the cost function4. The cost function is not provided in this problem5. We need to simplify the profit function using integers or decimals6. We need to find the number of roast beef sandwiches to maximize profit and the maximum weekly profit

STEP 2

The profit function is already given in a simplified form using integers or decimals. So, the answer to part a is $(x)=-0.001 x^{2}+2.4 x-530.00$

STEP 3

To find the number of roast beef sandwiches to maximize profit, we need to find the derivative of the profit function. This is because the maximum or minimum of a function occurs where its derivative is zero.
$'(x) = \frac{d}{dx}(-0.001 x^{2}+2. x-530.00)$

STEP 4

Calculate the derivative of the profit function.
$'(x) = -0.002x +2.4$

STEP 5

Set the derivative of the profit function equal to zero and solve for $x$ .
$-0.002x +2.4 =0$

STEP 6

olve for $x$ .
$x = \frac{2.4}{0.002}$

STEP 7

Calculate the value of $x$ .
$x = \frac{2.4}{0.002} =1200$

STEP 8

To ensure that this value of $x$ gives a maximum profit, we can take the second derivative of the profit function and substitute $x=1200$ . If the second derivative is negative, then $x=1200$ gives a maximum profit.
$''(x) = \frac{d^2}{dx^2}(-0.001 x^{2}+2.4 x-530.00)$

STEP 9

Calculate the second derivative of the profit function.
$''(x) = -.002$

STEP 10

The second derivative is negative, so $x=1200$ gives a maximum profit. Therefore, the store should make and sell1200 roast beef sandwiches each week to maximize profit.

STEP 11

To find the maximum weekly profit, substitute $x=1200$ into the profit function.
$(1200) = -0.001 \times1200^{}+.4 \times1200-530.00$

STEP 12

Calculate the maximum weekly profit.
$(1200) = -0.001 \times1200^{2}+2.4 \times1200-530.00 = \$770.00$ The store should make and sell1200 roast beef sandwiches each week to maximize profit, and the maximum weekly profit is $770.00.

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