라그랑주 승수법으로 주어진 함수가 제한 조건 하에서 갖는 최댓값과 최솟값이 무엇인지 구해보도록 물어봅시다.
[!user] User Use Lagrange multipliers to find the absolute maximum and minimum values of the function $f(x, y)=x^3+y^3-\frac{3}{2} x^2 y^2$ on the circle $x^2+y^2=5$
[!gpt] ChatGPT To use the Lagrange multiplier method to optimize a function subject to a constraint, we set up the following system of equations:
- $\nabla f = \lambda \nabla g$, where $f(x, y) = x^3 + y^3 - \frac{3}{2} x^2 y^2$ is the function to be optimized and $g(x, y) = x^2 + y^2 - 5$ is the constraint.
- $g(x, y) = 0$
Here, $\nabla f$ and $\nabla g$ are the gradients of $f$ and $g$, respectively, and $\lambda$ is the Lagrange multiplier.
The gradient of $f$ is: \(\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\)
Similarly, the gradient of $g$ is: \(\nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right)\)
Let’s start by computing the partial derivatives:
- $\frac{\partial f}{\partial x}$
- $\frac{\partial f}{\partial y}$
- $\frac{\partial g}{\partial x}$
- $\frac{\partial g}{\partial y}$
ChatGPT
The partial derivatives are:
- $\frac{\partial f}{\partial x} = 3x^2 - 3xy^2$
- $\frac{\partial f}{\partial y} = -3x^2y + 3y^2$
- $\frac{\partial g}{\partial x} = 2x$
- $\frac{\partial g}{\partial y} = 2y$
Now, we’ll set up the system of equations using the Lagrange multiplier method:
- $3x^2 - 3xy^2 = \lambda \times 2x$
- $-3x^2y + 3y^2 = \lambda \times 2y$
- $x^2 + y^2 = 5$
Let’s solve for $x$, $y$, and $\lambda$.
ChatGPT
We obtained the following solutions for $x$, $y$, and $\lambda$:
- $x = -2$, $y = 1$, $\lambda = -\frac{9}{2}$
- $x = 1$, $y = -2$, $\lambda = -\frac{9}{2}$
- $x = 0$, $y = -\sqrt{5}$, $\lambda = -\frac{3\sqrt{5}}{2}$
- $x = -\sqrt{5}$, $y = 0$, $\lambda = -\frac{3\sqrt{5}}{2}$
- $x = 0$, $y = \sqrt{5}$, $\lambda = \frac{3\sqrt{5}}{2}$
- $x = \sqrt{5}$, $y = 0$, $\lambda = \frac{3\sqrt{5}}{2}$
- $x = -\frac{\sqrt{10}}{2}$, $y = -\frac{\sqrt{10}}{2}$, $\lambda = -\frac{15}{4} - \frac{3\sqrt{10}}{4}$
- $x = \frac{\sqrt{10}}{2}$, $y = \frac{\sqrt{10}}{2}$, $\lambda = -\frac{15}{4} + \frac{3\sqrt{10}}{4}$
Now, we’ll evaluate the function $f(x, y)$ at each of these points to determine the absolute maximum and minimum values.
ChatGPT
The values of $f(x, y)$ at the solution points are:
- $-13$
- $-13$
- $-5\sqrt{5}$
- $-5\sqrt{5}$
- $5\sqrt{5}$
- $5\sqrt{5}$
- $-9.375 - \frac{5\sqrt{10}}{2}$
- $-9.375 + \frac{5\sqrt{10}}{2}$
From these values:
- The absolute maximum value of $f$ on the circle $x^2 + y^2 = 5$ is $5\sqrt{5}$.
- The absolute minimum value of $f$ on the circle $x^2 + y^2 = 5$ is $-9.375 - \frac{5\sqrt{10}}{2}$.
라그랑주 승수법으로 구한 답이 실제로 기하학적 상황과 일관적인지를 확인하기 위해 질문해봅시다.
[!user] User x+y+z/루트2 = 2와 x^2+(y^2 + z^2)/2 =1은 서로 접점 (1/2, 1, 1/루트2)에서 tangent한가? 영어로 답하라.
[!gpt] ChatGPT
To determine if the two surfaces are tangent at the point $(1/2, 1, 1/\sqrt{2})$, we’ll check if the normal vectors (gradients) of the surfaces at that point are parallel. If they are, the two surfaces are tangent to each other at that point.
The two given surfaces are:
- $x + y + \frac{z}{\sqrt{2}} = 2$
- $x^2 + \frac{y^2 + z^2}{2} = 1$
The gradient (normal vector) of the first surface at any point $(x, y, z)$ is the vector of its partial derivatives:
\[\nabla f_1 = \left( \frac{\partial f_1}{\partial x}, \frac{\partial f_1}{\partial y}, \frac{\partial f_1}{\partial z} \right)\]Similarly, the gradient of the second surface at any point is:
\[\nabla f_2 = \left( \frac{\partial f_2}{\partial x}, \frac{\partial f_2}{\partial y}, \frac{\partial f_2}{\partial z} \right)\]Where:
- $f_1(x, y, z) = x + y + \frac{z}{\sqrt{2}}$
- $f_2(x, y, z) = x^2 + \frac{y^2 + z^2}{2}$
Let’s start by computing these gradients.
ChatGPT
It seems that there was a technical issue. Let’s retry the computation.
We’ll compute the gradients (normal vectors) for the surfaces $f_1$ and $f_2$.
ChatGPT
I apologize for the inconvenience. We seem to be experiencing technical difficulties.
However, I can guide you through the process without direct computations:
For the first surface $x + y + \frac{z}{\sqrt{2}} = 2$, the gradient is straightforward: \(\nabla f_1 = \langle 1, 1, \frac{1}{\sqrt{2}} \rangle\)
For the second surface $x^2 + \frac{y^2 + z^2}{2} = 1$, the gradient is: \(\nabla f_2 = \left\langle 2x, y, z \right\rangle\)
To check if they are tangent at the point $(1/2, 1, 1/\sqrt{2})$, evaluate the gradient of $f_2$ at this point and check if the resulting vector is a scalar multiple of $\nabla f_1$.
If the vectors are scalar multiples of each other, then the two surfaces are tangent at the given point. Otherwise, they aren’t.
Given the difficulties, would you like to proceed with this approach, or would you prefer another method?
[!user] User yes
[!gpt] ChatGPT Alright, let’s evaluate the gradient $\nabla f_2$ at the point $(1/2, 1, 1/\sqrt{2})$ and see if it’s a scalar multiple of $\nabla f_1$.
Given: \(\nabla f_2 = \left\langle 2x, y, z \right\rangle\)
At $(1/2, 1, 1/\sqrt{2})$: \(\nabla f_2(1/2, 1, 1/\sqrt{2}) = \left\langle 2(1/2), 1, 1/\sqrt{2} \right\rangle = \left\langle 1, 1, 1/\sqrt{2} \right\rangle\)
Comparing the two vectors: \(\nabla f_1 = \left\langle 1, 1, 1/\sqrt{2} \right\rangle\) \(\nabla f_2(1/2, 1, 1/\sqrt{2}) = \left\langle 1, 1, 1/\sqrt{2} \right\rangle\)
Since $\nabla f_1$ and $\nabla f_2$ at the given point are identical, they are parallel and the two surfaces are tangent to each other at the point $(1/2, 1, 1/\sqrt{2})$.