Science:Math Exam Resources/Courses/MATH105/April 2013/Question 05 (b)
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) •
Question 05 (b) 

Use the method of Lagrange multipliers to find the maximum and minimum values of on the circle A solution that does not use the method of Lagrange multipliers will receive no credit, even if the answer is correct. 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! 
Hint 

Let be the constraint function. We thus need to solve the system of equations given by

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. Let be the constraint function. We thus need to solve the system of equations given by Plugging in our values gives Expanding the second equation gives the following three equations Substituting the first equation in the third equation gives which simplifies to Reducing gives . This tells us that and thus we can divide by it to get that .
Case 1: x = 0Plugging x = 0 into gives .
Case 2: x ≠ 0In this case we divide the equation above by x and obtain
Solving gives . Thus . Plugging these y values into gives .
Compare function values at all candidate pointsWe found the following six solutions: Plugging in each of the values gives
and thus the minimum value is 4 and the maximum value is 4. 