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Quadratic Word Problems (profit/gravity) A company sells widgets. The amount of profit, y, made by the company, is related tothe selling price of each widget, x, by the given equation. Using this equation, find out the maximum amount of profit the company can make, to the nearest dollar.y= -35x2 + 1458x – 8400

Sagot :

[tex]\begin{gathered} \text{The quadratic equation is,} \\ y=-35x^2+1458x-8400 \\ \text{for maxi}mum\text{ amount,} \\ \frac{dy}{dx}=0 \\ \frac{d(-35x^2+1458x-8400)}{dx}=0 \\ -70x+1458=0 \\ x=\frac{1458}{70} \\ x=20.82 \\ So,\text{ the maxi}mum\text{ profit,} \\ y=-35(20.82)^2+1458(20.82)-8400 \\ y=\text{ \$6772 (Profit)} \end{gathered}[/tex]

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