Three different solutions to one problem.
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?
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,
{ 8 x − 4 y = λ ⋅ 2 x − 4 x + 2 y = λ ⋅ 2 y x 2 + y 2 = 25 ⇒ { 4 x − 2 y = λ ⋅ x ( 1 ) 4 x − 2 y = λ ⋅ ( − 2 y ) ( 2 ) x 2 + y 2 = 25 ( 3 )
\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}
⎩ ⎨ ⎧ 8 x − 4 y = λ ⋅ 2 x − 4 x + 2 y = λ ⋅ 2 y x 2 + y 2 = 25 ⇒ ⎩ ⎨ ⎧ 4 x − 2 y = λ ⋅ x ( 1 ) 4 x − 2 y = λ ⋅ ( − 2 y ) ( 2 ) x 2 + y 2 = 25 ( 3 )
Then,
( 1 ) + ( 2 ) ⇒ λ ⋅ x = λ ⋅ ( − 2 y ) ⇒ λ ⋅ ( x − 2 y ) = 0 ⇒ λ = 0 or x = − 2 y .
(1)+(2)\Rightarrow \lambda\cdot x=\lambda\cdot (-2y)\Rightarrow \lambda\cdot (x-2y)=0\Rightarrow \lambda=0\text{ or } x=-2y.
( 1 ) + ( 2 ) ⇒ λ ⋅ x = λ ⋅ ( − 2 y ) ⇒ λ ⋅ ( x − 2 y ) = 0 ⇒ λ = 0 or x = − 2 y .
Sub $\lambda=0$ into (1), $4x-2y=0 \Rightarrow y=2x$.
Putting them together, $x=-2y$ or $y=2x$.
Case 1.
Sub $x=-2y$ into $(3) \Rightarrow (-2y)^2+y^2=25 \Rightarrow 5y^2=25\Rightarrow y^2=5$.
{ x = − 2 5 y = 5 or { x = 2 5 y = − 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}
{ x = − 2 5 y = 5 or { x = 2 5 y = − 5 case 2.
Sub $y=2x$ into $(3) \Rightarrow x^{2}+(2 x)^{2}=25 \Rightarrow 5 x^{2}=25 \Rightarrow x^{2}=5$.
{ x = 5 y = 2 5 or { x = − 5 y = − 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}
{ x = 5 y = 2 5 or { x = − 5 y = − 2 5
(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 5 − 5 ) 2 = ( − 5 5 ) 2 = 25 × 5 = 125.
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.
T ( x , y ) = 4 x 2 − 4 x y + y 2 = ( 2 x − y ) 2 = ( − 4 5 − 5 ) 2 = ( − 5 5 ) 2 = 25 × 5 = 125.
(2) Sub $x=2 \sqrt{5}, y=-\sqrt{5}$ into $T(x, y)$,
T ( x , y ) = ( 2 x − y ) 2 = ( 4 5 + 5 ) 2 = ( 5 5 ) 2 = 125.
T(x, y)=(2 x-y)^{2}=(4 \sqrt{5}+\sqrt{5})^{2}=(5 \sqrt{5})^{2}=125.
T ( x , y ) = ( 2 x − y ) 2 = ( 4 5 + 5 ) 2 = ( 5 5 ) 2 = 125.
(3) Sub $x=\sqrt{5}, y=2 \sqrt{5}$ into $T(x, y)$,
T ( x , y ) = ( 2 x − y ) 2 = 0.
T(x, y)=(2 x-y)^{2}=0.
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.
T(x, y)=(2 x-y)^{2}=0.
T ( x , y ) = ( 2 x − y ) 2 = 0.
Hence, the highest temperature is $125$, and the lowest one is $0$.
Recall:
from the equation $(1)$ and $(2)$, we get $x=-2 y$ or $y=2 x$.
Case 1.
Sub $x=-2 y$ into (3), $(-2 y)^{2}+y^{2}=25 \Rightarrow y^{2}=5$
T ( x , y ) = 4 x 2 − 4 x y + y 2 = ( 2 x − y ) 2 = ( 2 ⋅ ( − 2 y ) − y ) 2 = 25 y 2 = 25 × 5 = 125.
\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}
T ( x , y ) = 4 x 2 − 4 x y + y 2 = ( 2 x − y ) 2 = ( 2 ⋅ ( − 2 y ) − y ) 2 = 25 y 2 = 25 × 5 = 125. Case2.
Sub $y=2 x$ into (3), $x^{2}+(2 x)^{2}=25 \Rightarrow x^{2}=5$
T ( x , y ) = 4 x 2 − 4 x y + y 2 = ( 2 x − y ) 2 = ( y − y ) 2 = 0.
\begin{aligned}
T(x, y) &=4 x^{2}-4 x y+y^{2} \\\
&=(2 x-y)^{2} \\\
&=(y-y)^{2} \\\
&=0.
\end{aligned}
T ( x , y ) = 4 x 2 − 4 x y + y 2 = ( 2 x − y ) 2 = ( y − y ) 2 = 0. Hence, the highest temperature is $125$,
and the lowest one is $0$.
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
f x = λ g x f y = λ g y g ( x , y ) = 0
\begin{aligned}
f_{x}&=\lambda g_{x} \\\
f_{y}&=\lambda g_{y} \\\
g(x, y)&=0
\end{aligned}
f x f y g ( x , y ) = λ g x = λ g y = 0
for some constant $\lambda$, called a Lagrange multiplier.
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
{ x = 5 cos θ y = 5 sin θ
\begin{cases}
x=5 \cos \theta\\\
y=5 \sin \theta
\end{cases}
{ x = 5 cos θ y = 5 sin θ
and then sub into $(1)$,
T ( x , y ) = ( 2 x − y ) 2 = ( 10 cos θ − 5 sin θ ) 2 = ( 5 ⋅ ( 2 cos θ − sin θ ) ) 2 = 25 ⋅ ( 2 cos θ − sin θ ) 2 = 25 ⋅ ( 5 ⋅ cos ( θ + α ) ) 2 = 25 ⋅ 5 ⋅ cos 2 ( θ + α ) = 125 ⋅ cos 2 ( θ + α ) .
\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}
T ( x , y ) = ( 2 x − y ) 2 = ( 10 cos θ − 5 sin θ ) 2 = ( 5 ⋅ ( 2 cos θ − sin θ ) ) 2 = 25 ⋅ ( 2 cos θ − sin θ ) 2 = 25 ⋅ ( 5 ⋅ cos ( θ + α ) ) 2 = 25 ⋅ 5 ⋅ cos 2 ( θ + α ) = 125 ⋅ cos 2 ( θ + α ) . 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 $0$.
