Profit. Suppose that the monthly revenue and cost (in dollars) for x units of a product are $R = 200 x - \frac { x ^ { 2 } } { 10 } \text { and } C = 6000 + 10 x$ At what rate per month is the profit changing if the number of units produced and sold is 100 and is increasing at a rate of 10 units per month?
A) $\$ 18,800$ per month
B) $\$ 1700$ per month
C) $\$ 1800$ per month
D) $\$ - 100$ per month
E) $\$ 19,800$ per month

Question

Quizplus · Accepted Answer

Profit, $P$, is given by $P = R - C$. To find how the profit is changing, we differentiate $P$ with respect to time, $t$, to get $\frac{dP}{dt}$. Given $R = 200x - \frac{x^2}{10}$ and $C = 6000 + 10x$, we find $\frac{dR}{dt}$ and $\frac{dC}{dt}$, and use $x = 100$ and $\frac{dx}{dt} = 10$ units/month. The calculation yields $\frac{dP}{dt} = \$1700$ per month.

Profit Suppose That the Monthly Revenue and Cost (In Dollars) R=200x−x210 and C=6000+10xR = 200 x - \frac { x ^ { 2 } } { 10 } \text { and } C = 6000 + 10 xR=200x−10x2​ and C=6000+10x

Profit Suppose That the Monthly Revenue and Cost (In Dollars) $R = 200 x - \frac { x ^ { 2 } } { 10 } \text { and } C = 6000 + 10 x$