One Exercise about Lagrange Multiplier

Three different solutions to one problem.

Question

The temperature at a point $(x,y)$ on a metallic floor is $T(x,y)=4x^2-4xy+y^2$. Wearing thermocouples shoes that measures temperature, you walk along a circle of radius $5$ centered at the origin on the metallic floor. What are the highest and lowest temperatures measured by your shoes?

Solution

Lagrange Multiplier

The target: $T(x,y)=4x^2-4xy+y^2\Rightarrow T_x=8x-4y, T_y=-4x+2y$.

The constraint: $g(x,y)=x^2+y^2-25=0\Rightarrow g_x=2x, g_y=2y$.

By the method of Lagrange Multiplier, $$ \begin{cases} 8x-4y=\lambda\cdot 2x\\ -4x+2y=\lambda\cdot 2y\\ x^2+y^2=25 \end{cases} \Rightarrow \begin{cases} 4x-2y=\lambda\cdot x\quad (1)\\ 4x-2y=\lambda\cdot (-2y)\quad (2)\\ x^2+y^2=25 \quad (3) \end{cases} $$ Then, $$ (1)+(2)\Rightarrow \lambda\cdot x=\lambda\cdot (-2y)\Rightarrow \lambda\cdot (x-2y)=0\Rightarrow \lambda=0\text{ or } x=-2y. $$ Sub $\lambda=0$ into (1), $4x-2y=0 \Rightarrow y=2x$.

Putting them together, $x=-2y$ or $y=2x$.

Method 1

Case 1. Sub $x=-2y$ into $(3) \Rightarrow (-2y)^2+y^2=25 \Rightarrow 5y^2=25\Rightarrow y^2=5$. $$ \begin{cases} x=-2\sqrt{5}\\ y=\sqrt{5} \end{cases} \quad \text{or} \quad \begin{cases} x=2\sqrt{5}\\ y=-\sqrt{5} \end{cases} $$

case 2. Sub $y=2x$ into $(3) \Rightarrow x^{2}+(2 x)^{2}=25 \Rightarrow 5 x^{2}=25 \Rightarrow x^{2}=5$. $$ \begin{cases} x=\sqrt{5}\\ y=2\sqrt{5} \end{cases} \quad \text{or} \quad \begin{cases} x=-\sqrt{5}\\ y=-2\sqrt{5} \end{cases} $$ (1) Sub $x=-2 \sqrt{5}, y=\sqrt{5}$ into $T(x, y)$, $$ T(x, y)=4 x^{2}-4 x y+y^{2}=(2 x-y)^{2}=(-4 \sqrt{5}-\sqrt{5})^{2}=(-5 \sqrt{5})^{2}=25 \times 5=125. $$ (2) Sub $x=2 \sqrt{5}, y=-\sqrt{5}$ into $T(x, y)$, $$ T(x, y)=(2 x-y)^{2}=(4 \sqrt{5}+\sqrt{5})^{2}=(5 \sqrt{5})^{2}=125. $$ (3) Sub $x=\sqrt{5}, y=2 \sqrt{5}$ into $T(x, y)$, $$ T(x, y)=(2 x-y)^{2}=0. $$ (4) Sub $x=\sqrt{5}, y=-2 \sqrt{5}$ into $T(x, y)$ $$ T(x, y)=(2 x-y)^{2}=0. $$ Hence, the highest temperature is $125$, and the lowest one is $0$.

Method 2. Shortcut

Recall: from the equation $(1)$ and $(2)$, we get $x=-2 y$ or $y=2 x$.

Case 1. Sub $x=-2 y$ into (3), $(-2 y)^{2}+y^{2}=25 \Rightarrow y^{2}=5$

$$ \begin{aligned} T(x, y) &=4 x^{2}-4 x y+y^{2} \\ &=(2 x-y)^{2} \\ &=(2 \cdot(-2 y)-y)^{2} \\ &=25 y^{2}=25 \times 5=125 . \end{aligned} $$

Case2. Sub $y=2 x$ into (3), $x^{2}+(2 x)^{2}=25 \Rightarrow x^{2}=5$

$$ \begin{aligned} T(x, y) &=4 x^{2}-4 x y+y^{2} \\ &=(2 x-y)^{2} \\ &=(y-y)^{2} \\ &=0. \end{aligned} $$

Hence, the highest temperature is $125$, and the lowest one is $0$.

Remark

Result 1.8A

The maximum/minimum value of $f(x, y)$ subject to the constraint $g(x, y)=0$ occurs at a point $(x, y)$ that satisfies the following three equations $$ \begin{aligned} f_{x}&=\lambda g_{x} \\ f_{y}&=\lambda g_{y} \\ g(x, y)&=0 \end{aligned} $$ for some constant $\lambda$, called a Lagrange multiplier.

Parametric equations

The target: $T(x, y)=4 x^{2}-4 x y+y^{2} \quad (1)$.

The constraint: $g(x, y)=x^{2}+y^{2}-25=0 \quad (2)$.

The parametric equation for (2) is $$ \begin{cases} x=5 \cos \theta\\ y=5 \sin \theta \end{cases} $$ and then sub into $(1)$, $$ \begin{aligned} T(x, y)=(2 x-y)^{2} &=(10 \cos \theta-5 \sin \theta)^{2} \\ &=(5 \cdot(2 \cos \theta-\sin \theta))^{2}\\ &=25 \cdot(2 \cos \theta-\sin \theta)^{2} \\ &=25 \cdot(\sqrt{5} \cdot \cos (\theta+\alpha))^{2} \\ &=25 \cdot 5 \cdot \cos ^{2}(\theta+\alpha)\\ &=125 \cdot \cos ^{2}(\theta+\alpha). \end{aligned} $$

Note that $A\cos\theta-B\sin\theta=\sqrt{A^2+B^2}\cdot \cos (\theta+\alpha)$, where $\alpha=\tan^{-1}\left(\frac{B}{A}\right)$. Hence $2\cos\theta-\sin\theta=\sqrt{5}\cdot\cos(\theta+\alpha)$, $\alpha=\tan^{-1}(\frac{1}{2})$ in line $4$.

Note also $-1 \leqslant \cos (\theta+\alpha) \leqslant 1 \Rightarrow 0 \leqslant \cos ^{2}(\theta+\alpha) \leqslant 1 \Rightarrow 0 \leqslant T(x, y) \leqslant 125$.

Hence, the highest temperature is $125$, and the lowest one is $0$.