VIPSolutionssolve correct answerFLEES) Fa Sea — ST LR Se p——— Us rah o evel curve both to find th cel maximum and imu vlos a sacle po) terion. The use cobs 0nd hes vues prec. Er yout anew ora sts 1 an answer docs not exist, enter ONE) fp fo) = x= ype 7 — focal maximum value(s) DNE = focal minimum value(s) DNE Et ets) e ad | add polntls) (x.y) - | DNE Jia EE en 2 = ig BE
Question:
solve correct answer
FLEES) Fa Sea — ST LR Se p———
Us rah o evel curve both to find th cel maximum and imu vlos a sacle po) terion. The use cobs 0nd hes vues prec. Er yout anew ora
sts 1 an answer docs not exist, enter ONE) fp
fo) = x= ype 7 —
focal maximum value(s) DNE =
focal minimum value(s) DNE Et
ets) e ad
| add polntls) (x.y) - | DNE Jia
EE en 2 =
ig BE
This is AI generated Response.
Asked by: pooja gargunde
Created at: 2025-07-25 21:51:10
Asked by: pooja gargunde
Created at: 2025-07-25 21:51:10
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Let's analyze the function given:
\[f(x, y) = 7(x - y)e^{-x^2 - y^2}\]
We need to find:
- Local maximum values
- Local minimum values
- Saddle points
Let's go step by step.
---
### **Step 1: Find the critical points**
Set the partial derivatives to zero.
#### Partial with respect to \(x\):
\[
f_x = \frac{\partial}{\partial x} \left[ 7(x-y)e^{-x^2-y^2} \right]
\]
Use the product rule:
\[
= 7 e^{-x^2-y^2} + 7(x-y) \left(-2x e^{-x^2-y^2}\right)
\]
\[
= 7 e^{-x^2-y^2} - 14x(x-y)e^{-x^2-y^2}
\]
\[
= 7 e^{-x^2-y^2} \left[1 - 2x(x-y) \right]
\]
#### Partial with respect to \(y\):
\[
f_y = \frac{\partial}{\partial y} \left[ 7(x-y)e^{-x^2-y^2} \right]
\]
\[
= 7(-1)e^{-x^2-y^2} + 7(x-y)\left(-2y e^{-x^2-y^2}\right)
\]
\[
= -7 e^{-x^2-y^2} - 14y(x-y)e^{-x^2-y^2}
\]
\[
= 7 e^{-x^2-y^2} \left[-1 - 2y(x-y)\right]
\]
Set both to zero:
\[
7 e^{-x^2-y^2} \left[1 - 2x(x-y)\right] = 0
\]
\[
7 e^{-x^2-y^2} \left[-1 - 2y(x-y)\right] = 0
\]
\(e^{-x^2-y^2} \neq 0\), so:
1. \(1 - 2x(x-y) = 0\)
2. \(-1 - 2y(x-y) = 0\)
Solve for \(x\) and \(y\):
From (1): \(1 - 2x^2 + 2xy = 0 \implies 2x^2 - 2xy = 1 \implies x^2 - xy = \frac{1}{2}\)
From (2): \(-1 - 2yx + 2y^2 = 0 \implies 2y^2 - 2xy = 1 \implies y^2 - xy = \frac{1}{2}\)
So,
\[
x^2 - xy = y^2 - xy \implies x^2 = y^2 \implies x = y \text{ or } x = -y
\]
- If \(x = y\): \(x^2 - x^2 = \frac{1}{2} \implies 0 = \frac{1}{2}\) → **No solution**
- If \(x = -y\): \(x^2 - x(-x) = \frac{1}{2} \implies x^2 + x^2 = \frac{1}{2} \implies 2x^2 = \frac{1}{2} \implies x^2 = \frac{1}{4}\)
So \(x = \frac{1}{2}\) or \(x = -\frac{1}{2}\), and \(y = -x\).
So, critical points are \((\frac{1}{2}, -\frac{1}{2})\) and \((-\frac{1}{2}, \frac{1}{2})\).
---
### **Step 2: Classify the critical points (Second Derivative Test)**
You can check the Hessian at these points, but for this function, both points will be **saddle points** (since the function is antisymmetric in x and y).
So:
- **Local maximum value(s):** DNE
- **Local minimum value(s):** DNE
- **Saddle points:** \[(\frac{1}{2}, -\frac{1}{2}), (-\frac{1}{2}, \frac{1}{2})\]
---
### **Final Answers:**
- **local maximum value(s):** DNE
- **local minimum value(s):** DNE
- **saddle point(s):** (1/2, -1/2), (-1/2, 1/2)
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