# The cost function for a certain company is C=60x+300 and the revenue

The cost function for a certain company is
$C=60x+300$
and the revenue is given by
$R=100x-0.5{x}^{2}$
Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $\mathrm{}300$.
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Vaing1990
Step 1
$P\left(x\right)=R\left(x\right)-C\left(x\right)$
Where P = Profit Function, C = cost function, R = revenue function
Step 2
$R\left(x\right)-C\left(x\right)=300$
$⇒100x-0.5{x}^{2}-60x-300=300$
$⇒40x-0.5{x}^{2}-600=0$
$⇒40x-\frac{5}{10}{x}^{2}-600=0$
$⇒400x-5{x}^{2}-6000=0$
$⇒5{x}^{2}-400x+6000=0$
$⇒{x}^{2}-80x+1200=0$
$⇒{x}^{2}-60x-20x+1200=0$
$⇒x\left(x-60\right)-20\left(x-60\right)=0$
$⇒\left(x-60\right)\left(x-20\right)=0$
Rither, $x-60=0$
$⇒x=60$
or, $x-20=0$
$x=20$
$\therefore$ The two values of x with be 60, 20 that with create a profit of Rs 300