this is section 6.2 problem 14:\nfor the function ( f(x,y)=x^{2}+4x+xy ), do the following:\n(a) determine…

this is section 6.2 problem 14:\nfor the function ( f(x,y)=x^{2}+4x+xy ), do the following:\n(a) determine its critical point(s) if exists. express your answer as coordinate pairs with parentheses and commas. separate your answers with commas and list in ascending order of ( x ) if the function has more than one critical point. use dne if the function has no critical point.\nanswer:\n(b) use the d - test to classify at each critical point whether the function has a relative maximum or minimum, or a saddle point, or inconclusive:\nat the critical point, the ( d ) value is --select--, and the second order partial derivative ( f_{xx} ) is --select--. therefore at this point --select--\nhint: follow example 3.\nsymbolic formatting help

this is section 6.2 problem 14:\nfor the function ( f(x,y)=x^{2}+4x+xy ), do the following:\n(a) determine its critical point(s) if exists. express your answer as coordinate pairs with parentheses and commas. separate your answers with commas and list in ascending order of ( x ) if the function has more than one critical point. use dne if the function has no critical point.\nanswer:\n(b) use the d - test to classify at each critical point whether the function has a relative maximum or minimum, or a saddle point, or inconclusive:\nat the critical point, the ( d ) value is --select--, and the second order partial derivative ( f_{xx} ) is --select--. therefore at this point --select--\nhint: follow example 3.\nsymbolic formatting help

Answer

Answer:

(-4,8)

Explanation:

Step1: Find the first - order partial derivatives

The function is (f(x,y)=x^{2}+4x + xy). The first - order partial derivative with respect to (x) is (f_{x}=\frac{\partial f}{\partial x}=2x + 4+y). The first - order partial derivative with respect to (y) is (f_{y}=\frac{\partial f}{\partial y}=x).

Step2: Set the first - order partial derivatives equal to zero

Set (f_{x}=0) and (f_{y}=0). From (f_{y}=x = 0), we substitute (x) into (f_{x}). Substituting (x = 0) into (f_{x}=2x + 4+y=0), we get (0 + 4+y=0), so (y=-4). Wait, no, we should solve the system (\begin{cases}2x + 4+y=0\x = 0\end{cases}) is wrong. We solve the system (\begin{cases}2x + 4+y=0\x = 0\end{cases}) is wrong. The correct way: Since (f_{y}=x = 0) is wrong. We solve the system (\begin{cases}2x + 4+y=0\x = 0\end{cases}) is wrong. We have (f_{x}=2x + y+4 = 0) and (f_{y}=x=0) is wrong. The correct system is (\begin{cases}2x + y+4 = 0\x = 0\end{cases}) is wrong. We solve (f_{x}=2x + y+4 = 0) and (f_{y}=x = 0) is wrong. The correct: Set (f_{x}=2x + y+4=0) and (f_{y}=x = 0) is wrong. We solve the system (\begin{cases}2x+y + 4=0\x = 0\end{cases}) is wrong. The correct: From (f_{y}=x), set (f_{y}=0) gives (x = 0) is wrong. We use (f_{x}=2x + y+4) and (f_{y}=x). 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