Science:Math Exam Resources/Courses/MATH200/April 2012/Question 05 (a)

MATH200 April 2012

  •  Q1 (a)  •  Q1 (b)  •  Q2 (a)  •  Q2 (b)  •  Q3  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7  •  Q8  •  Q9  •  Q10  •  

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  •  April 2012  •  December 2011  •  

Question 05 (a)
Let $C$ be the intersection of the plane $\displaystyle {}x+y+z=2$ and the sphere $\displaystyle {}x^{2}+y^{2}+z^{2}=2.$ Use Lagrange multipliers to find the maximum value of $f(x,y,z)=z$ on $C$.

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Hint
Get the equations in this Lagrange multiplier problem by setting the partial derivatives of the objective function, $\Lambda$, to zero, where $\displaystyle \Lambda (x,y,z,\mu ,\nu )=f(x,y,z)+\mu (x+y+z-2)+\nu (x^{2}+y^{2}+z^{2}-2)$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We wish to maximize the value of $f(x,y,z)=z$ subject to the constraints $x+y+z=2,\quad x^{2}+y^{2}+z^{2}=2.$ So we use Lagrange multipliers. The objective function, $\Lambda$, is $\displaystyle \Lambda (x,y,z,\mu ,\nu )=f(x,y,z)+\mu (x+y+z-2)+\nu (x^{2}+y^{2}+z^{2}-2)$ Computing the partial derivatives of $\Lambda$ with respect to all the variables $x,y,z$ and the Lagrange multipliers $\mu ,\nu$ and setting them equal to zero gives: ${\begin{aligned}(1)\quad {\frac {\partial \Lambda }{\partial x}}&=\mu +2\nu x=0\\(2)\quad {\frac {\partial \Lambda }{\partial y}}&=\mu +2\nu y=0\\(3)\quad {\frac {\partial \Lambda }{\partial z}}&=1+\mu +2\nu z=0\\(4)\quad {\frac {\partial \Lambda }{\partial \mu }}&=x+y+z-2=0\\(5)\quad {\frac {\partial \Lambda }{\partial \nu }}&=x^{2}+y^{2}+z^{2}-2=0\end{aligned}}$ To begin with, we observe that $\nu$ can not be zero, because if it was, then equations (1) and (2) would force $\mu =0$, while equation (3) would force $\mu =-1$; a contradiction. Hence, we combine equations (1) and (2) to see that the maximum of $f$ occurs on the plane $y=x$: ${\begin{cases}\mu +2\nu x=0\\\mu +2\nu y=0\end{cases}}\quad \rightarrow \quad y=x$ Taking this result and using (4), we find $2x+z-2=0\quad \rightarrow \quad x={\frac {2-z}{2}}$ Taking these two results and using (5) we have an equation for z: $x^{2}+y^{2}+z^{2}-2=\left({\frac {2-z}{2}}\right)^{2}+\left({\frac {2-z}{2}}\right)^{2}+z^{2}-2=0$ Multiplying this equation by 4 and expanding all the quadratic terms we get: $\displaystyle 8-8z+6z^{2}-8=6z^{2}-8z=0$ Solving this equation for $z$ we get $\displaystyle z=0,\ {\frac {4}{3}}.$ Taking the larger of the two results we obtain the maximum of $f(x,y,z)=z$ subject to the given constraints which is $z=4/3$.

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