Science:Math Exam Resources/Courses/MATH105/April 2017/Question 03

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MATH105 April 2017

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Question 03
Use the method of Lagrange Multipliers to find the maximum value of the utility function $U=f(x,y)=16x^{\frac {1}{4}}y^{\frac {3}{4}}$ , subject to the constraint $G(x,y)=50x+100y-500,000=0$ , where $x\geq 0$ and $y\geq 0$ . You do not need to justify your answer. Note that a solution that does not use the method of Lagrange Multipliers will receive no credit, even if the answer is correct.

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Hint
The method of Lagrange multipliers states that a maximum in the interior of the domain must occur at a place where the following equations are satisfied: ${\begin{aligned}{\frac {\partial f}{\partial x}}&=\lambda {\frac {\partial {G}}{\partial {x}}}\\{\frac {\partial f}{\partial y}}&=\lambda {\frac {\partial {G}}{\partial {y}}}\\G(x,y)&=0\end{aligned}}$ . This is a system of three equations in x, y, and $\lambda$ . Solve this system by eliminating the variable $\lambda$ .

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Solution
The method of Lagrange multipliers states that a maximum in the interior of the domain must occur at a place where the following equations are satisfied: ${\begin{aligned}{\frac {\partial f}{\partial x}}&=\lambda {\frac {\partial {G}}{\partial {x}}}\\{\frac {\partial f}{\partial y}}&=\lambda {\frac {\partial {G}}{\partial {y}}}\\G(x,y)&=0\end{aligned}}$ . We now compute the derivatives: ${\begin{aligned}{\frac {\partial f}{\partial x}}&=4x^{-3/4}y^{3/4}\\{\frac {\partial G}{\partial x}}&=50\\{\frac {\partial f}{\partial y}}&=12x^{1/4}y^{-1/4}\\{\frac {\partial G}{\partial y}}&=100\end{aligned}}$ . So we need to solve the system ${\begin{aligned}4x^{-3/4}y^{3/4}&=50\lambda \\12x^{1/4}y^{-1/4}&=100\lambda \\50x+100y&=500000\end{aligned}}$ . If we multiply the first equation by 2, we get $8x^{-3/4}y^{3/4}=100\lambda$ . Substituting this into the second equation gives $12x^{1/4}y^{-1/4}=8x^{-3/4}y^{3/4}$ . Finally, we multiply both sides by $x^{3/4}y^{1/4}$ to arrive at the equation $12x=8y$ , and solve for y to get $y={\frac {3}{2}}x.$ We now plug into the third equation: $50x+100\cdot {\frac {3}{2}}x=500000$ . This simplifies to $200x=500000$ which comes out to $x=2500.$ From this, it is clear that $y=3750$ Giving a value of $f(x,y)=16(2500)^{1/4}(3750)^{3/4}$ . This can be further simplified to $f(x,y)=20000\cdot 54^{1/4}.$ According to the problem statement, this is all we need to do - the problem stated that we do not need to provide justification. We will provide a justification here nonetheless. The family of points satisfying G(x,y) = 0 form a line in the plane. The family of such points with $x\geq 0$ and with $y\geq 0$ are a line segment with endpoints on the y-axis and x-axis. Because the maximum must either occur at a critical point or at an endpoint of the domain, and because f(x,y) = 0 if either x or y is zero, it follows that f achieves its maximum at the critical point (2500, 3750). Answer: $\color {blue}20000{\sqrt[{4}]{54}}$