Science:Math Exam Resources/Courses/MATH105/April 2010/Question 04

MATH105 April 2010

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[hide] Question 04
The production function for a firm is $\displaystyle f(x,y)=5x^{2/3}y^{1/3},$ where $x$ and $y$ are the number of units of labour and capital utilized respectively. Suppose that labour costs $108 per unit and capital costs $2 per unit and that the firm decides to produce 600 units of goods. Use Lagrange multipliers to determine the amounts of labour and capital that should be utilized in order to minimize cost. You need not show that your solution minimizes the cost.

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[show] Hint
What are the objective function and the constraint function?

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[show] Solution
We are given that labour costs $108 per unit and capital costs $2 per unit, and we want to minimize cost. Thus the objective function is $\displaystyle C(x,y)=108x+2y$ . The firm wants to produce 600 units, so using the production function we want $\displaystyle 5x^{2/3}y^{1/3}=600$ . Thus, after dividing both sides by 5, the constraint function is $\displaystyle g(x,y)=x^{2/3}y^{1/3}-120$ , and the constraint is g(x, y) = 0. The Lagrange multiplier method says that the gradients of the objective function and the constraint function should be proportional, or $\nabla C(x,y)=\lambda \nabla g(x,y)$ . The gradient of the objective function is $\nabla C(x,y)=\langle 108,2\rangle$ . and the gradient of the constraint function is $\nabla g(x,y)=\langle {\frac {2}{3}}x^{-1/3}y^{1/3},{\frac {1}{3}}x^{2/3}y^{-2/3}\rangle$ Then we have three equations that must be satisfied $108=\lambda {\frac {2}{3}}x^{-1/3}y^{1/3}$ , $2=\lambda {\frac {1}{3}}x^{2/3}y^{-2/3}$ , $\displaystyle x^{2/3}y^{1/3}-120=0$ . Solving the second equation for $\lambda$ gives $\lambda =6x^{-2/3}y^{2/3}$ , and substituting into the first equation to eliminate $\lambda$ gives $108=6x^{-2/3}y^{2/3}{\frac {2}{3}}x^{-1/3}y^{1/3}$ . Simplifying, we have $\displaystyle 108=4x^{-1}y$ , and then we solve for y to get $\displaystyle y=27x$ . Now substituting into the third equation, the constraint equation, gives $\displaystyle x^{2/3}(27x)^{1/3}-120=0$ . Simplifying, we have $\displaystyle 3x=120$ , and thus $\displaystyle x=40$ . We then have that $\displaystyle {}y=27x=27(40)=1080.$ Therefore the cost is minimized when using 40 units of labour and 1080 units of capital.