Complete Question:
A soft-drink vendor at a popular beach analyzes his sales records, and finds that if he sells x cans of soda pop in one day, his profit (in dollars) is given by P(x) = -0.001x² + 3x - 1800.
a. What is his maximum profit per day?
b. How many cans must be sold in order to obtain the maximum profit?
Answer:
a. $450
b. 1500 cans
Step-by-step explanation:
Given the following quadratic function;
P(x) = -0.001x² + 3x - 1800 ......equation 1
a. To find his maximum profit per day;
Since P(x) is a quadratic equation, P(x) would be maximum when [tex] x = \frac {-b}{2a} [/tex]
Note : the standard form of a quadratic equation is ax² + bx + c = 0 ......equation 2
Comparing eqn 1 and eqn 2, we have;
a = -0.001, b = 3 and c = -1800
Now, we determine the maximum profit;
[tex] x = \frac {-b}{2a} [/tex]
Substituting the values, we have;
[tex] x = \frac {-3}{2*(-0.001)} [/tex]
Cancelling out the negative signs, we have;
[tex] x = \frac {3}{2*0.001} [/tex]
[tex] x = \frac {3}{0.002} [/tex]
x at maximum = 1500
Substituting the value of "x" into equation 1;
P(1500) = -0.001 * 1500² + 3(1500) - 1800
P(1500) = -0.001 * 2250000 + 4500 - 1800
P(1500) = -2250 + 2700
P(1500) = $450
b. Therefore, the soft-drink vendor must sell 1500 cans in order to obtain the maximum profit.
