The cost of controlling emissions at a firm goes up rapidly as the amount of emissions reduced goes up. Here is a possible model. $C ( x , y ) = 6,000 + 100 x ^ { 2 } + 25 y ^ { 2 }$
Where x is the reduction in sulfur emissions, y is the reduction in lead emissions (in pounds of pollutant per day), and C is the daily cost to the firm (in dollars) of this reduction. Government clean-air subsidies amount to $600 per pound of sulfur and $50 per pound of lead removed. How many pounds of pollutant should the firm remove each day to minimize the net cost (cost minus subsidy)

A)3.5 lb of sulfur and 2 lb of lead
B) 6 lb of sulfur and 1 lb of lead
C) 3 lb of sulfur and 1 lb of lead
D) 1.9 lb of sulfur and 3.5 lb of lead
E) 1 lb of sulfur and 3 lb of lead

Question

The cost of controlling emissions at a firm goes up rapidly as the amount of emissions reduced goes up. Here is a possible model.   $C ( x , y ) = 6,000 + 100 x ^ { 2 } + 25 y ^ { 2 }$  
Where x is the reduction in sulfur emissions, y is the reduction in lead emissions (in pounds of pollutant per day), and C is the daily cost to the firm (in dollars) of this reduction. Government clean-air subsidies amount to $600 per pound of sulfur and $50 per pound of lead removed. How many pounds of pollutant should the firm remove each day to minimize the net cost (cost minus subsidy) 
 
A)3.5 lb of sulfur and 2 lb of lead
B) 6 lb of sulfur and 1 lb of lead
C) 3 lb of sulfur and 1 lb of lead
D) 1.9 lb of sulfur and 3.5 lb of lead
E) 1 lb of sulfur and 3 lb of lead

Quizplus · Accepted Answer

To minimize the net cost, we need to consider both the cost of controlling emissions and the subsidies received. The net cost function can be written as:$$Net \ Cost = C(x, y) - (600x + 50y)$$Substituting the given cost function:$$Net \ Cost = 6000 + 100x^2 + 25y^2 - (600x + 50y)$$To find the minimum, we take the partial derivatives with respect to x and y, set them to zero, and solve for x and y.$$\frac{\partial}{\partial x}(6000 + 100x^2 + 25y^2 - 600x - 50y) = 200x - 600 = 0$$$$\frac{\partial}{\partial y}(6000 + 100x^2 + 25y^2 - 600x - 50y) = 50y - 50 = 0$$Solving these equations gives:$$200x - 600 = 0 \Rightarrow x = 3$$$$50y - 50 = 0 \Rightarrow y = 1$$Therefore, the firm should remove 3 pounds of sulfur and 1 pound of lead each day to minimize the net cost.

The Cost of Controlling Emissions at a Firm Goes Up C(x,y)=6,000+100x2+25y2C ( x , y ) = 6,000 + 100 x ^ { 2 } + 25 y ^ { 2 }C(x,y)=6,000+100x2+25y2

The Cost of Controlling Emissions at a Firm Goes Up $C ( x , y ) = 6,000 + 100 x ^ { 2 } + 25 y ^ { 2 }$