The cost of controlling emissions at a firm goes up rapidly as the amount of emi

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) = 3,000 + 200x2 + 100- 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 900 per pound of sulfur and $500 per pound of lead removed. How many pounds of pollutant should the firm remove each day to minimize net cost (cost minus subsidy)? lbs of sulfur removed per day lbs of lead removed per day

Explanation / Answer

given

The daily cost is given as:

C(x,y) = 3000 + 200x^2 + 100y^2

Subtract the subsidy to get the net daily cost. This is the function we wish to minimize:

N(x,y) = 3000 + 200x^2 + 100y^2 - 900x - 500y

In any problem involving minimizing or maximizing a function, we want to find the function's critical points; i.e. take the derivative and solve for the derivative equaling zero. Since x and y are independent of each other, y drops out like a constant when you take the derivative with respect to x, and vice versa.

dN/dx = 400x - 900
dN/dy = 200y - 500

dN/dx = 0
400x - 900 = 0
x = 9/4

dN/dy = 0
200y - 500 = 0
y = 5/2

So the total cost has only one critical point with respect to x (x = 9/4) and only one critical point with respect to y (y = 5/2). Use the second derivative test to verify that these critical points are minima.

dN/dx = 400x - 900
d2N/dx2 = 400
d2N/dx2 > 0 therefore x = 9/4 is a minimum

dN/dy = 200y - 500
d2N/dy2 = 200
d2N/dy2 > 0 therefore y = 5/2 is a minimum

The firm should remove 9/4 of a pound of sulfur and 5/2 pound of lead per day.

x=9/4

y=5/2

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