the cost of controlling emissions at a firm is given by c(q)=5000 + 120q², where q is the reduction in…

the cost of controlling emissions at a firm is given by c(q)=5000 + 120q², where q is the reduction in emissions (in pounds of pollutants per day) and c is the daily cost to the firm (in dollars) of this reduction. government clean - air subsidies amount to $840 per pound of pollutant removed. how many pounds of pollutants should the firm remove each day in order to minimize net cost (cost minus subsidy)? pounds of pollutant

the cost of controlling emissions at a firm is given by c(q)=5000 + 120q², where q is the reduction in emissions (in pounds of pollutants per day) and c is the daily cost to the firm (in dollars) of this reduction. government clean - air subsidies amount to $840 per pound of pollutant removed. how many pounds of pollutants should the firm remove each day in order to minimize net cost (cost minus subsidy)? pounds of pollutant

Answer

Explanation:

Step1: Define the net - cost function

Let the net - cost function $N(q)$ be the cost function $C(q)$ minus the subsidy. The cost function is $C(q)=5000 + 120q^{2}$, and the subsidy is $840q$ per pound of pollutant removed. So, $N(q)=5000 + 120q^{2}-840q$.

Step2: Find the derivative of the net - cost function

The derivative of $N(q)$ with respect to $q$ is $N^\prime(q)=\frac{d}{dq}(5000 + 120q^{2}-840q)$. Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $N^\prime(q)=240q-840$.

Step3: Set the derivative equal to zero to find critical points

Set $N^\prime(q) = 0$. So, $240q-840 = 0$. Add 840 to both sides: $240q=840$. Then solve for $q$: $q=\frac{840}{240}=\frac{7}{2}=3.5$.

Step4: Check the second - derivative to confirm it's a minimum

The second - derivative $N^{\prime\prime}(q)=\frac{d}{dq}(240q - 840)=240>0$. Since the second - derivative is positive, $q = 3.5$ is a point of minimum of the net - cost function.

Answer:

3.5 pounds of pollutant