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Problem of the Week #34 - November 19th, 2012

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Chris L T521

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Jan 26, 2012
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Thanks to those who participated in last week's POTW!! Here's this week's problem!

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Problem: A company incurs costs of $C(x,y)=5x^2+2xy+3y^2+800$ (in thousands of dollars) when it produces $x$ thousand units of one product and $y$ thousand units of another. Its production capacity is such that $x+y=39$. At what production levels will the company's costs be minimized? What will be the corresponding total cost?

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Suggestion:

Solve this using the method of Lagrange multipliers.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's problem was correctly answered by Ackbach, Bacterius, BAdhi, MarkFL, and Sudharaka. MarkFL showed it using Lagrange multipliers (and regular algebra), Sudharaka showed it using Lagrange multipliers as well, and everyone else used single variable calculus. You can find MarkFL's answer below (for the Lagrange multiplier and algebra solution):

We are given the objective function:

$\displaystyle C(x,y)=5x^2+2xy+3y^2+800$


subject to the constraint:


$\displaystyle g(x,y)=x+y-39=0$


Using the theorem of Lagrange, we obtain the system:


(1) $\displaystyle 10x+2y=\lambda$


(2) $\displaystyle 2x+6y=\lambda$


These equations imply:


$\displaystyle y=2x$


Substituting for $\displaystyle y$ into the constraint, there results:


$\displaystyle x=13\,\therefore\,y=26$


Hence, the minimum cost in thousands of dollars at the given production level is:


$\displaystyle C(13,26)=4349$


Alternate solution without calculus:


Substitute for $\displaystyle y$ using the constraint into the objective function to obtain:


$\displaystyle C(x)=6x^2-156x+5363$


The minimum point, i.e., the vertex, lies on the axis of symmetry, given by:


$\displaystyle x=-\frac{-156}{12}=13$


Hence, the minimum cost in thousands of dollars at the given production level is:


$\displaystyle C(13)=4349$

Here's Bacterius' solution using single variable calculus:

As $x + y = 39$, there is only one degree of freedom, so the cost can be rewritten in a single variable:

$C(x, y) = C(x) = 5x^2 + 2x(39 - x) + 3(39 - x)^2 + 800 = 6x^2 - 156x + 5363$


Which is minimized by:


$C'(x) = 12x - 156 \implies x = 13 \iff y = 26$


Hence the production levels which minimize the company's costs are $x = 13$ and $y = 26$, with a total cost of $C(13, 26) = 4349$ (thousands of dollars).
 
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