2.9E: Optimization of Functions of Several Variables (Exercises)
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13.8: Optimization of Functions of Several Variables
Finding Critical Points
In exercises 1 - 5, find all critical points.
1) \( f(x,y)=1+x^2+y^2\)
- Answer:
- \( (0,0)\)
2) \( f(x, y) = 1 - (x -2)^2 + (y+3)^2\)
3) \( f(x,y)=(3x−2)^2+(y−4)^2\)
- Answer:
- \( \left(\frac{2}{3},4\right)\)
4) \( f(x,y)=x^4+y^4−16xy\)
- Answer:
- \( (0,0), \quad (-2,-2), \quad (2,2)\)
5) \( f(x,y)=15x^3−3xy+15y^3\)
- Answer:
- \( (0,0), \quad \left(\frac{1}{15},\frac{1}{15}\right)\)
Finding Extrema & the Second Partials Test
In exercises 6 - 9, find the critical points of the function and test for extrema or saddle points by using algebraic techniques (completing the square) or by examining the form of the equation. Where possible, verify your results using the Second Partials Test.
6) \( f(x,y)=-\sqrt{x^2+y^2}\)
- Answer:
-
Crit. pts.: \( (0, 0) \)
Extrema: \( f\) has a relative maximum of \(0\) at \( (0, 0)\).
To justify this, consider the fact that the square root function cannot give a negative value, so this function cannot possibly return a positive value. Since it's value is \(0\) at the critical point \( (0, 0)\), we know it must be the function's absolute maximum value.
7) \( f(x,y)=−x^2−5y^2+8x−10y−13\)
- Answer:
-
Crit. pts.: \( (4, -1) \)
Extrema: \( f\) has a relative maximum of \(8\) at \( (4,−1)\).
To justify this, we complete the square on this function, being careful to factor out the coefficient of the squared terms before completing the square.
\[\begin{align*} f(x, y) &= −x^2−5y^2+8x−10y−13 \\ &= −(x^2-8x\quad\quad)−5(y^2+2y\quad\quad)−13 \\ &= −(x^2-8x+16)−5(y^2+2y+1)−13+16+5 \\ &= -(x-4)^2 -5(y+1)^2+8\end{align*}\]
Note that this quadratic polynomial function takes the form \( z = -(x^2 + y^2)\), so we can see that it will have a relative (and, in fact, absolute) maximum at its vertex (the critical point \( (4, -1) \)). We can also argue that since we are subtracting squared terms from 8, we cannot possibly obtain a function value larger than 8, and since we do get a value of 8 at the critical point \( (4, -1) \), we know it will be the absolute maximum value of this function.
8) \( f(x,y)=x^2+y^2+2x−6y+6\)
9) \( f(x,y)=\sqrt{x^2+y^2}+1\)
- Answer:
-
Crit. Pts.: \( (0, 0) \)
Extrema: \( f\) has a relative minimum of \(1\) at \( (0,0)\).
To justify this, consider the fact that the square root function cannot give a negative value, so this function cannot possibly return a value less than \(1\). Since it's value is \(1\) at the critical point \( (0, 0)\), we know \(1\) must be the function's absolute minimum value.
In exercises 10 - 34, identify any critical points and use the Second Partials Test to determine the behavior of the function at each critical point, whether there is a maximum, minimum, saddle point, or none of these. If the Second Partials Test fails, determine the function's behavior at the point using another method and justify your answer clearly.
10) \( f(x,y)=−x^3+4xy−2y^2+1\)
11) \( f(x,y)=x^2y^2\)
- Answer:
-
Crit. pts.: All points on the lines \( x = 0 \) and \( y = 0\) are critical points of this function.
Extrema: The Second Partials Test fails.
Since \( x^2y^2>0\) for all \( x\) and \( y\) different from zero, and \( x^2y^2=0\) when either \( x\) or \( y\) equals zero (or both), then the absolute minimum of \(0\) occurs at all points on the \(x\)- or \(y\)-axes, that is, for all points on the lines \( x = 0 \) and \( y = 0\).
12) \( f(x,y)=x^2−6x+y^2+4y−8\)
13) \( f(x,y)=2xy+3x+4y\)
- Answer:
-
Crit. pts.: \( \left(−2,−\frac{3}{2}\right) \)
Exrema: \(f\) has a saddle point at \( \left(−2,−\frac{3}{2},−6\right) \).
14) \( f(x,y)=8xy(x+y)+7\)
15) \( f(x,y)=x^2+4xy+y^2\)
- Answer:
-
Crit. pts.: \( (0,0) \)
Exrema: \(f\) has a saddle point at \( (0,0,0)\).
16) \( f(x,y)=x^3+y^3−300x−75y−3\)
17) \( f(x,y)=9−x^4y^4\)
- Answer:
-
Crit. pts.: All points on the lines \( x = 0 \) and \( y = 0\) are critical points of this function.
Extrema: The Second Partials Test fails.
Since the term \( -x^4y^4<0\) for all \( x\) and \( y\) different from zero, and \( -x^4y^4=0\) when either \( x\) or \( y\) equals zero (or both), then this function cannot attain a value greater than \(9\) anywhere, but is \(9\) at the critical points. Thus \(f\) has an absolute maximum of \(9\) at all points on the \(x\)- or \(y\)-axes, that is, for all points on the lines \( x = 0 \) and \( y = 0\).
18) \( f(x,y)=x^2+10xy+y^2\)
- Answer:
-
Crit. pts.: \( (0,0) \)
Extrema: \(f\) has a saddle point at \( (0,0,0)\).
19) \(f(x,y) = x^4 + y^2 + 2xy + 3\)
- Answer:
-
Crit. pts.: \( (0,0), \quad \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right), \quad \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right) \)
Extrema: \(f\) has a saddle point at \( (0, 0, 3) \),
\(f\) has a local minimum of \( 2.75\) at the point \( \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) \).
\(f\) has a local minimum of \( 2.75\) at the point \( \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right) \).
20) \( f(x,y)=7x^2y+9xy^2\)
21) \( f(x,y)=3x^2−2xy+y^2−8y\)
- Answer:
-
Crit. pts.: \( (2,6) \)
Extrema: \(f\) has a relative minimum of \( -24\) located at \( (2,6)\).
22) \( f(x,y)=3x^2+2xy+y^2\)
23) \( f(x,y)=y^2+xy+3y+2x+3\)
- Answer:
-
Crit. pts.: \( (1,−2) \)
Extrema: \(f\) has a saddle point at \( (1,−2,1)\).
24) \( f(x,y)=x^2+xy+y^2−3x\)
25) \( f(x,y)=x^2+2y^2−x^2y\)
- Answer:
-
Crit. pts.: \( (0,0), \quad (-2,1), \quad (2,1)\)
Extrema: \(f\) has a relative minimum of \(0\) at \( (0,0) \) and saddle points at \( (2,1,2)\) and \( (−2,1,2)\).
26) \( f(x,y)=x^2+y−e^y\)
27) \( f(x,y)=e^{−(x^2+y^2+2x)}\)
- Answer:
-
Crit. pts.: \( (-1,0) \)
Extrema: \(f\) has a relative maximum of \( e \) located at \( (-1,0)\).
See this problem illustrated in CalcPlot3D .
28) \( f(x,y)=x^2+xy+y^2−x−y+1\)
29) \(f(x,y) = x^2y(9 - x + y)\)
- Answer:
-
Crit. pts.: \( \left(\frac{9}{2},-\frac{9}{4}\right), \quad (9,0)\), and all points on the line \(x = 0\)
Extrema: \(f\) has a saddle point at \( (9,0,0)\) and a relative minimum of \(-102.515625\) at \( \left(\frac{9}{2},-\frac{9}{4}\right)\).
At the critical points on the line \(x = 0\), \(f\) has neither relative extrema nor saddle points, but they do represent a sort of trough on the surface.
30) \( f(x,y)=−x^2−5y^2+10x−30y−62\)
31) \( f(x,y)=120x+120y−xy−x^2−y^2\)
- Answer:
-
Crit. pts.: \( (40,40) \)
Extrema: \(f\) has a relative maximum of \( 4800 \) located at \( (40,40)\).
32) \( f(x,y)=2x^2+2xy+y^2+2x−3\)
33) \( f(x,y)=x^2+x−3xy+y^3−5\)
- Answer:
-
Crit. pts.: \( \left(\frac{1}{4},\frac{1}{2}\right)\) and \((1, 1) \)
Extrema: \(f\) has a saddle point at \( \left(\frac{1}{4},\frac{1}{2}, -\frac{79}{16}\right)\) and a relative minimum of \( -5 \) at \( (1,1)\).
34) \( f(x,y)=2xye^{−x^2−y^2}\)
In exercises 35 - 37, determine the extreme values and the saddle points. Use a CAS to graph the function.
35) [T] \( f(x,y)=ye^x−e^y\)
- Answer:
-
A saddle point is located at \( (0,0,-1).\)
36) [T] \( f(x,y)=x\sin(y)\)
37) [T] \( f(x,y)=\sin(x)\sin(y),\quad x∈(0,2π),\quad y∈(0,2π)\)
- Answer:
-
There is a saddle point at \( (π,π),\) local maxima at \( \left(\frac{π}{2},\frac{π}{2}\right)\) and \( \left(\frac{3π}{2},\frac{3π}{2}\right)\), and local minima at \( \left(\frac{π}{2},\frac{3π}{2}\right)\) and \( \left(\frac{3π}{2},\frac{π}{2}\right)\).
Contributors
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Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .
- Paul Seeburger (Monroe Community College) created problems 19 and 29, and added dynamic figures for problems 27 and 35.