MATH SOLVE

2 months ago

Q:
# Mary repairs microwaves. her revenue is modeled by the function r(h)=20+30h for every h hours she spends repairing microwaves. her overhead cost is modeled by the function c(h)=10h2−80 . after how many hours does she break even?

Accepted Solution

A:

The correct answer is:

5 hours.

Explanation:

To find the break-even point, we set the two functions equal to each other:

20+30h = 10h²-80

Since this is a quadratic equation, we want it set equal to 0 to solve. First we will subtract 20 from each side:

20+30h-20 = 10h²-80-2030h = 10h²-100

Subtract 30h from each side:30h-30h = 10h²-100-30h0 = 10h²-30h-100

We will use the quadratic formula to solve this:

[tex] x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} [/tex]

Instead of x, we have h. Our values for a, b and c are:a = 10b = -30c = -100

This gives us:[tex] h=\frac{--30\pm \sqrt{(-30)^2-4(10)(-100)}}{2(10)}

\\

\\=\frac{30\pm \sqrt{900--4000}}{20}

\\

\\=\frac{30\pm \sqrt{900+4000}}{20}

\\

\\=\frac{30\pm \sqrt{4900}}{20}

\\

\\=\frac{30\pm 70}{20}=\frac{30-70}{20}\text{ or }\frac{30+70}{20}

\\

\\=\frac{-40}{20}\text{ or }\frac{100}{20}=-2\text{ or }5 [/tex]

Since a negative number of time makes no sense, we use h = 5.

5 hours.

Explanation:

To find the break-even point, we set the two functions equal to each other:

20+30h = 10h²-80

Since this is a quadratic equation, we want it set equal to 0 to solve. First we will subtract 20 from each side:

20+30h-20 = 10h²-80-2030h = 10h²-100

Subtract 30h from each side:30h-30h = 10h²-100-30h0 = 10h²-30h-100

We will use the quadratic formula to solve this:

[tex] x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} [/tex]

Instead of x, we have h. Our values for a, b and c are:a = 10b = -30c = -100

This gives us:[tex] h=\frac{--30\pm \sqrt{(-30)^2-4(10)(-100)}}{2(10)}

\\

\\=\frac{30\pm \sqrt{900--4000}}{20}

\\

\\=\frac{30\pm \sqrt{900+4000}}{20}

\\

\\=\frac{30\pm \sqrt{4900}}{20}

\\

\\=\frac{30\pm 70}{20}=\frac{30-70}{20}\text{ or }\frac{30+70}{20}

\\

\\=\frac{-40}{20}\text{ or }\frac{100}{20}=-2\text{ or }5 [/tex]

Since a negative number of time makes no sense, we use h = 5.