To make the desired​ profit, the smallest amount you can charge for each box is ​$

Question

Your class is selling boxes of flower seeds as a fundraiser. The total profit p depends on the amount x that your class charges for each box of seeds. The equation p=−0.5x^2+36x−150 models the profit of the fundraiser.​ What's the smallest​amount, in​ dollars, that you can charge and make a profit of at least ​$408​?

(Round to the nearest cent as​ needed.)

Answer 1

Hello, Gabriela,

We can answer this by either graphing or solving using the quadratic formula. We want the lowest value of p, the price per box of seeds, that will generate at least $408 in profit, p.

A. Graph

If the equation is graphed, the total profit will be the y-axis, and the x axis is the price per box of seeds. It is a parabola. Draw a line through y = $408 (for all values of x).. The point of intersection with the original equation closest to the origin will the minimum price that will produce $408 in profit. A higher price will generate greater profits to a maximum before profits begin falling as the price increases. The point at which y=408 crosses the graph again is the price at which profits will start falling below $408 [likely because the customers don't like the ripoff]. My graph shows these two points to be (22.6, 408) and (49.4, 408).

B. Quadratic Formula

A more accurate way to find the price is to use the quadratic formula. Since we want a profit of at least $408, we can use that value in the profit equation and solve for x.

$408 = −0.5x² + 36x−150

-0.5x² + 36x - 558 = 0

I get values of x of 22.58 and 49.42, same as with the graph, thankfully. These are the more accurate, since graphing requires some interpretation.

Bob

Answer 2

p = −0.5x^2+36x−150 ≥ 408; -0.5x² + 36x - 558 ≥ 0 or x² - 72x + 1116 ≤ 0;

To solve this inequality we solve equation x² - 72x + 1116 = 0; x = 36 ± √36² - 1116 = 36 ± 6√5.

So x = 22.58 or x = 49.42; so, solution of inequality is interval [22.58, 49.42].

x =$ 22.58 per box is minimum prise.