Harold Kessler

## Answered question

2021-12-16

To calculate: The break-even point for
$P\left(x\right)=-1100+120x-{x}^{2}$

### Answer & Explanation

Suhadolahbb

Beginner2021-12-17Added 32 answers

Step 1
Given Information:
Profit function, $P\left(x\right)=-1100+120x-{x}^{2}$
Formula Used:
Break-even point is the point, when profit is zero or revenue is equal to the cost.
$\text{Profit}=\text{Total revenue}-\text{Total cost}$
Zero product property:
The General form of quadratic equation is $a{x}^{2}+bx+c=0$
Where a, b, c are constants and $a\ne 0$
For real numbers a and b. $ab=0$ if and only if $a=0$ or $b=0$ or both.
Step 2
Calculation:
To find the break-even point, put revenue is equal to the cost.
$P\left(x\right)=0$
$-1100+120x-{x}^{2}=0$
${x}^{2}-110x-10x+1100=0$
Take the common terms outside the parenthesis
$x\left(x-110\right)-10\left(x-110\right)=0$
$\left(x-10\right)\left(x-110\right)=0$
Use the zero product property,
$x-10=0$ or $x-110=0$
$x=10$ or $x=110$
It is given that $x\le 100$
Hence, the break-even point is 10 units.

Esta Hurtado

Beginner2021-12-18Added 39 answers

Step 1
$P\left(x\right)=-1100+120x-{x}^{2}$
Quadratic polynomial can be factored using the transformation
$a{x}^{2}+bx+c=a\left(x-{x}_{1}\right)\left(x-{x}_{2}\right)$
where ${x}_{1}$ and ${x}_{2}$ are the solutions of the quadratic equation
$a{x}^{2}+bx+c=0$
$-{x}^{2}+120x-1100=0$
All equations of the form $a{x}^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$x=\frac{-120±\sqrt{{120}^{2}-4\left(-1\right)\left(-1100\right)}}{2\left(-1\right)}$
Square 120
$x=\frac{-120±\sqrt{14400-4\left(-1\right)\left(-1100\right)}}{2\left(-1\right)}$
Multiply -4 times -1
$x=\frac{-120±\sqrt{14400+4\left(-1100\right)}}{2\left(-1\right)}$
Multiply 4 times -1100
$x=\frac{-120±\sqrt{14400-4400}}{2\left(-1\right)}$
Add 14400 to -4400
$x=\frac{-120±\sqrt{10000}}{2\left(-1\right)}$
Take the square root of 10000
$x=\frac{-120±100}{2\left(-1\right)}$
Multiply 2 times -1
$x=\frac{-120±100}{-2}$
Now solve the equation $x=\frac{-120±100}{-2}$ when $±$ is plus
Add -120 to 100
$x=\frac{-20}{-2}$
Divide -20 by -2
$x=10$
Now solve the equation $x=\frac{-120±100}{-2}$ when $±$ is minus
Subtract 100 from -120
$x=\frac{-220}{-2}$
Divide -220 by -2
$x=110$
Factro the original expression using $a{x}^{2}+bx+c=a\left(x-{x}_{1}\right)\left(x-{x}_{2}\right)$. Substitute 10 for ${x}_{1}$ and 110 for ${x}_{2}$
$-{x}^{2}+120x-1100=-\left(x-10\right)\left(x-110\right)$

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