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Multivariable Calculus UAF MATH F253X

Dr. Latrice Bowman

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Section Maximum and Minimum Values

Worksheet DC-4 Reading

All of the readings for this course come from the OpenStax Calculus Volume 3 textbook, which is available for free online at https://openstax.org/details/books/calculus-volume-3. You can read the textbook online or download a PDF version. The textbook is also available in print from various retailers if you prefer a physical copy.
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For this lesson, you should read sections 4.7 Maxima/Minima Problems. You should read through the sections carefully, making sure to understand the definitions, examples, and key ideas. You should also work through the exercises at the end of each section to test your understanding and practice applying the concepts. Some of the exercises at the end of the section are assigned in the WP assignments.
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Worksheet DC-4 Videos

Maxima and Minima
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Worksheet DC-4 Written Practice

INFORMATION ABOUT ALL WRITTEN PRACTICE (WP) ASSIGNMENTS:
  • All of the WP problems come from our online textbook: https://openstax.org/details/books/calculus-volume-3
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  • Complete solutions to all WP assignments can be found in Canvas within the section Module.
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  • Links to the text and PDFs of the text can be found in the Canvas module under the Reading and Information.
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THE WP ASSIGNMENTS ARE DESIGNED TO PREPARE YOU FOR THE QUIZZES AND ASSESSMENTS
  • Work on every problem on every assignment.
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  • If you get stuck or do not understand a solution, ask questions.
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  • For each problem, try it once or twice before looking at solutions or asking for help.
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  • Mark the problems you did correctly on the first try.
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  • Highlight problems where you got help (using solutions, videos, a tutor, etc) -These are the problem types that you will need more practice on to be ready for the assessments.
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  • Understand that just copying the solutions instead of working through the problems will greatly reduce your chances of success in this class.
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WP8
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Maxima-Minima Problems: Complete problems 313, 319, 331, 345, 352, 354
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Worksheet DC-4 Sample Outcomes

For the quiz on this outcome you would be given 40 minutes to complete 1 problem. Here are samples of the types of problems you will encounter:
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1.

Given \(f(x,y)=2-x^4+2x^2-y^2\text{,}\) find and classify all critical points.
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Hint.
First find the partial derivatives and set them equal to zero. Use the second derivative test to classify each critical point.
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Answer.
(0,0,2) is a saddle point. (1,0,3) and (-1,0,3) are local maxima.
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Solution.
First find the first-order partials:
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\(\frac{\partial f}{\partial x} = -4x^3 + 4x\)
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\(\frac{\partial f}{\partial y} = -2y\)
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Setting these equal to zero:
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\(-4x^3 + 4x = 0\)
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\(-2y = 0\)
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Solving these equations gives the critical points: (0,0), (1,0), and (-1,0).
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Next, find the second-order partials:
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\(\frac{\partial^2 f}{\partial x^2} = -12x^2 + 4\)
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\(\frac{\partial^2 f}{\partial y^2} = -2\)
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\(\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 0\)
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Now, apply the second derivative test:
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\(D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2\)
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At (0,0):
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\(D = (4)(-2) - (0)^2 = -8\)
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Since \(D \lt 0\text{,}\) (0,0) is a saddle point.
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At (1,0):
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\(D = (-8)(-2) - (0)^2 = 16\)
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Since \(D \gt 0\) and \(\frac{\partial^2 f}{\partial x^2} = -8 \lt 0\text{,}\) (1,0) is a local maximum.
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At (-1,0):
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\(D = (-8)(-2) - (0)^2 = 16\)
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Since \(D \gt 0\) and \(\frac{\partial^2 f}{\partial x^2} = -8 \lt 0\text{,}\) (-1,0) is also a local maximum.
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Since the function value at (0,0) is 2, and the function value at (1,0) and (-1,0) is 3, we can conclude that (0,0,2) is a saddle point and (1,0,3) and (-1,0,3) are local maxima.
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2.

Find three positive numbers whose sum is 80 and whose product is a maximum.
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Hint.
find the critical points of the function \(f(x,y,z) = xyz\) and use the sum to rewrite this as a function of two variables.
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Answer.
The three positive numbers are all equal to \(\frac{80}{3}\text{.}\)
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Solution.
First let three positive numbers, \(x, y, z\text{.}\) Since their sum is 80, we have \(x + y + z = 80\text{.}\) We want to maximize \(f(x,y,z) = xyz\text{.}\) Using the constraint, we can write \(z = 80 - x - y\) and substitute this into the function:
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\(f(x,y) = xy(80 - x - y) = 80xy - x^2y - xy^2\)
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Now we find the critical points of this function of two variables.
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Find the first-order partials:
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\(\frac{\partial f}{\partial x} = 80y - 2xy - y^2\)
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\(\frac {\partial f}{\partial y} = 80x - x^2 - 2xy\)
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Setting these equal to zero:
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\(80y - 2xy - y^2 = 0\)
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\(80x - x^2 - 2xy = 0\)
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From the first equation, we can factor out y:
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\(y(80 - 2x - y) = 0\)
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Since y is positive, we have \(80 - 2x - y = 0\text{,}\) which gives us \(y = 80 - 2x\text{.}\)
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Substituting this into the second equation:
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\(80x - x^2 - 2x(80 - 2x) = 0\)
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\(80x - x^2 - 160x + 4x^2 = 0\)
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\(3x^2 - 80x = 0\)
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\(x(3x - 80) = 0\)
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Since x is positive, we have \(3x - 80 = 0\text{,}\) which gives us \(x = \frac{80}{3}\text{.}\)
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Substituting this back into the equation for y:
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\(y = 80 - 2\left(\frac{80}{3}\right) = 80 - \frac{160}{3} = \frac{240 - 160}{3} = \frac{80}{3}\)
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Substituting x and y back into the equation for z:
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\(z = 80 - \frac{80}{3} - \frac{80}{3} = \frac{240 - 80 - 80}{3} = \frac{80}{3}\)
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So the critical point is at \(\left(\frac{80}{3}, \frac{80}{3}, \frac{80}{3}\right)\text{.}\)
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We can use the second derivative test to classify this critical point. First, we find the second-order partials:
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\(\frac{\partial^2 f}{\partial x^2} = -2y\)
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\(\frac{\partial^2 f}{\partial y^2} = -2x\)
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\(\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 80 - 2x - 2y\)
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Now, we can apply the second derivative test:
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\(D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2\)
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At the critical point:
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\(D = (-2y)(-2x) - (80 - 2x - 2y)^2\)
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\(D = 4xy - (80 - 2x - 2y)^2\)
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\(D = 4\left(\frac{80}{3}\right)\left(\frac{80}{3}\right) - (80 - 2\left(\frac{80}{3}\right) - 2\left(\frac{80}{3}\right))^2\)
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\(D = \frac{25600}{9} - (80 - \frac{160}{3} - \frac{160}{3})^2\)
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\(D = \frac{25600}{9} - ( - \frac{80}{3})^2\)
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\(D = \frac{25600}{9} - \frac{6400}{9}\)
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\(D = \frac{19200}{9}\)
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Since \(D \gt 0\) and \(\frac{\partial^2 f}{\partial x^2} = -2y \lt 0\text{,}\) the critical point is a local maximum.
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Since the function value at this critical point is \(f\left(\frac{80}{3}, \frac{80}{3}, \frac{80}{3}\right) = \left(\frac{80}{3}\right)\left(\frac{80}{3}\right)\left(\frac{80}{3}\right) = \frac{512000}{27}\text{,}\) we can conclude that the three positive numbers that maximize the product are all equal to \(\frac{80}{3}\text{.}\)
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3.

Find the shortest distance from the point (2,1,-1) to the plane \(x+y-z=1\text{.}\)
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Hint.
Use the square of the distance formula from a point to a plane to set up the function and find the critical points.
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Answer.
The shortest distance is \(\sqrt{3}\) units.
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Solution.
First we use the square of the distance formula from a point to a plane.
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So the function becomes: \(f(x)=(x-2)^2+(y-1)^2+(z+1)^2\)
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We can use the plane equation to express z in terms of x and y: \(z = x + y - 1\text{.}\) Substituting this into the distance function gives us:
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\(f(x,y) = (x-2)^2 + (y-1)^2 + (x + y - 1 + 1)^2\)
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\(f(x,y) = (x-2)^2 + (y-1)^2 + (x + y)^2\)
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\(f(x,y) = 2x^2 + 2y^2 + 2xy - 4x - 2y + 5\)
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Now we find the critical points of this function of two variables.
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Find the first-order partials:
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\(\frac{\partial f}{\partial x} = 4x + 2y - 4\)
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\(\frac{\partial f}{\partial y} = 4y + 2x - 2\)
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Setting these equal to zero:
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\(4x + 2y - 4 = 0\)
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\(4y + 2x - 2 = 0\)
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From the first equation, we can express y in terms of x:
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\(2y = 4 - 4x\)
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\(y = 2 - 2x\)
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Substituting this into the second equation:
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\(4(2 - 2x) + 2x - 2 = 0\)
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\(8 - 8x + 2x - 2 = 0\)
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\(-6x + 6 = 0\)
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\(6x = 6\)
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\(x = 1\)
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Substituting this back into the equation for y:
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\(y = 2 - 2(1) = 0\)
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Substituting x and y back into the equation for z:
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\(z = 1 + 0 - 1 = 0\)
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So the critical point is at (1,0,0).
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Now we can use the second derivative test to classify this critical point. First, we find the second-order partials:
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\(\frac{\partial^2 f}{\partial x^2} = 4\)
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\(\frac{\partial^2 f}{\partial y^2} = 4\)
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\(\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 2\)
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Now, we can apply the second derivative test:
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\(D = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2\)
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At the critical point:
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\(D = (4)(4) - (2)^2 = 16 - 4 = 12\)
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Since \(D \gt 0\) and \(\frac{\partial^2 f}{\partial x^2} = 4 \gt 0\text{,}\) the critical point is a local minimum.
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Since the function value at this critical point is \(f(1,0) = 2(1)^2 + 2(0)^2 + 2(1)(0) - 4(1) - 2(0) + 5 = 3\text{,}\) the shortest distance from the point (2,1,-1) to the plane \(x+y-z=1\) is \(\sqrt{3}\) units.
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