word problem: solve 300 = -5x^2 + 100x + 20

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LaryssaLacerda

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The profits P of a small real estate company depend on the number of weekly listings, x, and are described by the function P(x)=-5x^2+10x+20.

a. How many weekly listings would result in a profit of $300.

b. What number of weekly listings would result in a maximum profit? What would that profit be?

c. Is there a number of weekly listings that would result in the company actually losing money? If so, determine the number of listings at which the company would begin losing money.
 
I can't seem to figure out if I have to do factoring.. I am completely lost.. Can you please help me.. Thank-you sooo much.

300=-5x^2+100x+20
300-20=-5x^2+100x+20-20
280=-5x^2+100x
280-280=-5x^2+100x-280
0=-5x^2+100x-280
0=(-5x^2+100x-280)/(-5)
0= x^2-20x+56
 
Hello Laryssa:

I thought something was amiss. You have a typo in your listing. The coefficient of x is actually 100, not 10?.

Set your quadratic to 300 and solve.

\(\displaystyle \L\\-5x^{2}+100x-280=0\)

Use the quadratic formula to solve.

There will be 2 points where this is satisified.


For part b, the max profit:

Use \(\displaystyle \L\\x=\frac{-b}{2a}\) to find the x value(number of listings) where the max occurs. This formula gives you the x-coordinate of the vertex.

Find that and plug it into your quadratic to find the corresponding y value(the max profit).

For part c, set your quadratic equal to 0 and solve for x. You will have two values. One will be negative, so it can be disregarded. Any x value higher than the positive result will give a negative profit.
 
-5x^2+100x-280=0
(-5x^2+100x-280)/(-5)=0
x^2-20x+56=0

I get the quadratic part for part a but I just don't get what you have to do beyond this point x^2-20x+56.. I don't know if factoring is needed of what happens.

Maybe I'm just not going through the right step..
 
Soroban answered for you. You've never seen the quadratic formula?.
 
Find the zeroes of the profit function. Then find where the profit function is negative.

Eliz.
 
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