Pricing In The Cost Of Pollution

Calculus Level 3

Under the Kyoto Protocol, each country is assigned a quota on the amount of greenhouse gases which it emits a year. In turn, these countries set quotas on the emissions produced by local businesses and organizations. Each operator has an allowance of carbon credits, and may choose to purchase additional carbon credits as required.

Caterpillar Inc. runs an industrial facility which produces tractors. They know that if they produce T T tractors a month, their net profit from production would be ( − T 2 + 50 T − 200 ) ( - T^2 + 50T - 200) thousand dollars. However, if they produce more than 10 tractors, then each additional tractor will cost them ten thousand dollars in carbon credits.

What is the optimal number of tractors which Caterpillar should produce, in order to maximize their profit?

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50 10 25 20

Chung Gene Keun by Chung Gene Keun , 42, USA

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2 solutions

Steven Perkins
Mar 22, 2014

Since we're subtracting 10 thousand for each tractor above 10, we can adjust the equation to be:

− T 2 + 40 T − 100 -T^{2} + 40T - 100

The constant term was changed to give the same value as the original equation with 10 tractors. But the constant term doesn't matter in optimizing the profit based on number of tractors produced.

You can now use calculus, or simply plug in the multiple choice answers. The maximum of 300 thousand is reached at:

20 \boxed{20}

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Hee Kon Kim
Mar 22, 2014

If T is less than 10, the maximum profit will be 300 when T=10. If T is more than 10, the profit function will be -T^2+40T-200 and the maximum profit is 400 when T=20 as the derivative of the function equals to zero at that point. Thus, the maximum profit is achieved when T=20

2 Helpful 0 Interesting 0 Brilliant 0 Confused

let it produce (t+10) tractors. now, let profit be a function of t as P(t). this is a new t, not the old one. maximize P(t)= - (10+t)^2 + 50(10+t) - 200 - 10t
{the extra 10t is bein subtracted for carbon emissions of the t tractors purchased after the first 10 tractor} expanding, get P(t) = - t^2 + 20t + 200. hence P ' (t) = - 2t + 20, which is zero at t= 10. hence, maximized. therefore, total no. of tractors purchsed = t + 10 = 20.

Ashwin Anand - 7 years, 2 months ago

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