Questions: Question 29 A concession stand at the Tennis Center sells a hamburger/drink combination dinner for 7. The profit, y (in dollars), can be approximated by y=-0.001 x^2+3.6 x-400 where x is the number of dinners prepared. (a) Find the number of dinners that should be prepared to maximize profit. (b) What is the maximum profit?

Question 29

A concession stand at the Tennis Center sells a hamburger/drink combination dinner for 7. The profit, y (in dollars), can be approximated by y=-0.001 x^2+3.6 x-400 where x is the number of dinners prepared.
(a) Find the number of dinners that should be prepared to maximize profit.
(b) What is the maximum profit?

Transcript text: Question 29 A concession stand at the Tennis Center sells a hamburger/drink combination dinner for $7. The profit, $y$ (in dollars), can be approximated by $y=-0.001 x^{2}+3.6 x-400$ where $x$ is the number of dinners prepared. (a) Find the number of dinners that should be prepared to maximize profit. (b) What is the maximum profit?

Solution

Solution Steps

To solve this problem, we need to find the vertex of the quadratic function representing the profit. The vertex form of a quadratic equation $ y = ax^2 + bx + c $ gives the maximum or minimum point. Since the coefficient of $ x^2 $ is negative, the parabola opens downwards, and the vertex will give the maximum profit. The x-coordinate of the vertex can be found using the formula $ x = -\frac{b}{2a} $.

Step 1: Find the Number of Dinners to Maximize Profit

To find the number of dinners $ x $ that should be prepared to maximize profit, we use the vertex formula for the quadratic function $ y = -0.001x^2 + 3.6x - 400 $. The x-coordinate of the vertex is given by:

\[ x = -\frac{b}{2a} = -\frac{3.6}{2 \times -0.001} = 1800 \]

Step 2: Calculate the Maximum Profit

Next, we substitute $ x = 1800 $ back into the profit equation to find the maximum profit $ y $:

\[ y = -0.001(1800)^2 + 3.6(1800) - 400 \] Calculating this gives:

\[ y = 2840 \]