fambrini wrote:At store A, the profit from the sale of x units of a certain product is given by the formula P = -50x² + 600x. What is the maximum profit that the store can have from the sale of that product?
A) 6
B) 12
C) 600
D) 1800
E) 3600
OA: D
Since we need to determine the maximum value of x, it may be easiest to input each answer choice into our given equation of -50x² + 600x to determine whether we get an integer value for x. Variable x must be an integer because it represents a number of "units," and we cannot have a fractional unit.
Let's start with answer choice E, the largest potential amount of profit.
-50x² + 600x = 3,600
Dividing the entire equation by -50 gives us:
x² - 12x = -72
x² - 12x + 72 = 0
Since there are not any integer values of x that multiply to 72 and sum to -12, we cannot solve for x, in which x is an integer. Thus, 3,600 cannot be the profit.
Next we can test answer choice D.
-50x² + 600x = 1,800
Dividing the entire equation by -50 gives us:
x² - 12x = -36
x² - 12x + 36 = 0
(x - 6)(x - 6) = 0
Thus, x = 6. Since 6 is a possible value of x, 1,800 can be the maximum possible profit.
Alternate Solution:
Notice the given formula P = -50x² + 600x is a quadratic function. Furthermore, the graph of this function will be a parabola opening downward and the vertex of this parabola will be the maximum point of the graph. Therefore, we can determine the maximum profit by finding the vertex of the parabola.
Recall that x = -b/(2a) is the formula to find the x-value of the vertex; therefore the number of units that maximizes the profit is:
x = -600/[2(-50)] = -600/-100 = 6 and thus the maximum profit is P = -50(6)^2 + 600(6) = -1800 + 3600 = 1800.
Answer:
D
