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The revenue function is y₁ = –4x² + 400x, the cost function is y₁ = x² – 100x + 8000, and the break-even number of units is 20 or 80.
What is a quadratic equation?
It is a polynomial that is equal to zero. Polynomial of variable power 2, 1, and 0 terms are there. Any equation having one term in which the power of the variable is a maximum of 2 then it is called a quadratic equation.
Jua Kali Products Ltd has been in operation for the last 10 years.
Its annual revenue and cost functions take the form of quadratic functions.
The following data was obtained from the records of the company.
Year Unit Sold Revenue Cost
2017 5 1900 7525
2018 10 3600 7100
2019 15 5100 6725
We know that the quadratic equation is given as
[tex]\rm y = ax^2 + bx + c[/tex]
Let y₁ be the revenue function, y₂ be the cost function and x be the unis sold.
Then the revenue function will be
1900 = 25a + 5b + c ...i
3600 = 100a + 10b + c ...ii
5100 = 225a + 15b + c ...iii
From equations (i), (ii), and (iii), we have
a = –4, b = 400, and c = 0
Then the revenue function will be
y₁ = –4x² + 400x
Similarly, the cost function will be
7525 = 25a + 5b + c ...1
7100 = 100a + 10b + c ...2
6725 = 225a + 15b + c ...3
From equations 1, 2, and 3, we have
a = 1, b = –100, and c = 8000
Then the cost function will be
y₁ = x² – 100x + 8000
For the break-even units, the cost function and the revenue function will be equal. Then we have
[tex]\begin{aligned} x^2 -100x + 8000 &= -4x^2 + 400x\\\\5x^2 -500x + 8000 &= 0\\\\x^2 - 100x + 1600 &= 0\\\\x^2 - 80 x - 20x + 1600 &= 0\\\\x(x-80) - 20 (x-80) &= 0\\\\(x-80)(x-20) &= 0\\\\x &= 20, 80 \end{aligned}[/tex]
More about the quadratic equation link is given below.
https://brainly.com/question/2263981