// Numbas version: finer_feedback_settings {"name": "Find the stationary point of a function of two variables (2)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["rawstatval", "lmin", "q1", "dvalue", "neither", "b", "a", "c", "lmax", "p1", "d", "f", "y", "x", "type", "desc"], "name": "Find the stationary point of a function of two variables (2)", "tags": ["1603", "Calculus", "calculus", "checked2015", "critical point", "derivative", "determinant", "differentiation", "Differentiation", "function of 2 variables", "functions of 2 variables", "functions of two variables", "Hessian", "MAS1603", "maximum", "minimum", "partial derivatives", "partial differentiation", "saddle point", "stationary points", "stationary points of functions of two variables", "tested1", "type of critical point"], "preamble": {"css": "", "js": ""}, "advice": "

a) The $(x,y)$ coordinates of the critical point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$:

\\[\\begin{eqnarray*} \\frac{\\partial f}{\\partial x} &=&0\\\\ \\\\ \\frac{\\partial f}{\\partial y}&=&0 \\end{eqnarray*} \\]

In this case you get two equations to solve for $x$ and $y$:

\\[\\begin{eqnarray*} \\simplify[std]{{3*a}x^2+{2*b}x*y}&=&0\\\\ \\\\ \\simplify[std]{{b}x^2+{2*c}y+{d}}&=&0 \\end{eqnarray*} \\]
The first equation factorises as \\[\\simplify{x*({3*a}x+{2*b}y)=0}\\] and we find that either:

\\[x=0,\\text{ or } \\simplify[std]{y={-3*a}x/{2*b}}\\]

On substituting $x=0$ into the second equation we get $y=\\simplify[std]{{-d}/2}$.

Hence $(0,\\simplify[std]{{-d}/2})$ is a critical point and it is the only critical point where $x=0$.

On substituting these values for $x$ and $y$ into $f(x,y)$ we get:

\\[f\\left(0,\\simplify[std]{{-d}/2}\\right) = \\simplify[std]{{rawstatval}}.\\]

b) The Hessian at a point $(x,y)$ is:

\\[\\left(\\begin{array}{cc}\\simplify[std]{{6*a}x+{2*b}y}&\\simplify[std]{{2*b}x}\\\\ \\simplify[std]{{2*b}x}&\\var{2*c}\\end{array}\\right)\\]

Evaluated at the critical point $(0,\\simplify[std]{{-d}/2})$ we obtain:

\\[\\left(\\begin{array}{cc}\\simplify[std]{{-d*b}}&0\\\\ 0&\\var{2*c}\\end{array}\\right)\\]

The determinant is $D=\\var{dvalue}$.

We see that on looking at the information given by Steps above we have:

$\\var{desc}$

", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"stepsPenalty": 0, "prompt": "

$a=\\;\\;\\;$[[0]]. Input as a fraction or an integer and not as a decimal.

Input the value of $f(x,y)$ at the critical point $(0,a)$:

$f(0,a)=\\;\\;$[[1]](Input to 2 decimal places).

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Input answer as a fraction or an integer, not a decimal

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Input answer as a fraction or an integer, not a decimal

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The $(x,y)$ coordinates of the stationary point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$

\\[\\begin{eqnarray*} \\frac{\\partial f}{\\partial x} &=&0\\\\ \\\\ \\frac{\\partial f}{\\partial y} &=&0 \\end{eqnarray*} \\]

In this case you get two equations to solve for $x$ and $y$

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "information"}], "scripts": {}, "marks": 0, "showCorrectAnswer": true, "type": "gapfill"}, {"stepsPenalty": 0, "prompt": "

Input the Hessian of the function evaluated at the critical point found above:

$H=\\;$[[0]]

Hence choose the type of the critical point;

[[1]]

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Minimum

", "

Maximum

", "

Saddle

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The Hessian is defined by:

\\[H=\\;\\left(\\begin{array}{cc} \\frac{\\partial^2 f}{\\partial x^2}& \\frac{\\partial^2 f}{\\partial x \\partial y}\\\\
\\frac{\\partial^2 f}{\\partial y \\partial x}&\\frac{\\partial^2 f}{\\partial y^2}\\end{array}\\right)\\]

Evaluate this at the point found above.

Fihd the determinant $D=\\det(H)$ and then the following tells us what type of critical point we have:

a) $D \\gt 0$ and $\\displaystyle \\frac{\\partial^2 f}{\\partial x^2}\\gt 0$ gives a local minimum.

b) $D \\gt 0$ and $\\displaystyle \\frac{\\partial^2 f}{\\partial x^2}\\lt 0$ gives a local maximum.

c) $D \\lt 0$ gives a saddle point

d) $D=0$ not known - needs further work.

In the following question find $a$ such that $(0,a)$ is a critical point of the function:

\\[f(x,y)=\\simplify[std]{{a}*x^3+{b}*x^2*y+{c}*y^2+{d}y+{f}}\\]

Also find the type of this critical point.

", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": "dvalue<>0"}, "variables": {"q1": {"definition": "random(-3..3)", "templateType": "anything", "group": "Ungrouped variables", "name": "q1", "description": ""}, "dvalue": {"definition": "-2*b*c*d", "templateType": "anything", "group": "Ungrouped variables", "name": "dvalue", "description": ""}, "c": {"definition": "random(-5..5 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "c", "description": ""}, "p1": {"definition": "random(-3..3)", "templateType": "anything", "group": "Ungrouped variables", "name": "p1", "description": ""}, "d": {"definition": "random(-5..5 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "d", "description": ""}, "f": {"definition": "random(-5..5)", "templateType": "anything", "group": "Ungrouped variables", "name": "f", "description": ""}, "type": {"definition": "switch(dvalue >0 and d*b<0,[1,0,0],dvalue>0 and d*b>0,[0,1,0],[0,0,1])", "templateType": "anything", "group": "Ungrouped variables", "name": "type", "description": ""}, "rawstatval": {"definition": "a*x^3+b*x^2*y+c*y^2+d*y+f", "templateType": "anything", "group": "Ungrouped variables", "name": "rawstatval", "description": ""}, "lmax": {"definition": "'Local maximum'", "templateType": "anything", "group": "Ungrouped variables", "name": "lmax", "description": ""}, "b": {"definition": "random(-5..5 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "a": {"definition": "random(1..5)", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}, "y": {"definition": "-d/2", "templateType": "anything", "group": "Ungrouped variables", "name": "y", "description": ""}, "x": {"definition": "0", "templateType": "anything", "group": "Ungrouped variables", "name": "x", "description": ""}, "neither": {"definition": "'Saddle point'", "templateType": "anything", "group": "Ungrouped variables", "name": "neither", "description": ""}, "lmin": {"definition": "'Local minimum'", "templateType": "anything", "group": "Ungrouped variables", "name": "lmin", "description": ""}, "desc": {"definition": "switch(type[0]=1,'$\\\\displaystyle D\\\\gt 0,\\\\;\\\\;\\\\frac{\\\\partial^2 f}{\\\\partial x^2}\\\\gt 0\\\\Rightarrow \\\\text{ local minimum}.$',\n type[1]=1,'$\\\\displaystyle D\\\\gt 0,\\\\;\\\\;\\\\frac{\\\\partial^2 f}{\\\\partial x^2}\\\\lt 0\\\\Rightarrow \\\\text{ local maximum}.$',\n '$\\\\displaystyle D \\\\lt 0 \\\\Rightarrow \\\\text{ saddle point}$.')", "templateType": "anything", "group": "Ungrouped variables", "name": "desc", "description": ""}}, "metadata": {"notes": "

10/07/2012:

Added tags.

Question appears to be working correctly.

24/12/2012:

Calculations checked, OK. Added tested1 tag.

12/1/2016

This copy asks for type of critical point using the Hessian.

", "description": "

Find the critical point $(0,a)$ of the function: $f(x,y)=ax^3+bx^2y+cy^2+dy+f$ and find its type using the test given by the Hessian matrix.

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