Quadratic equation to find max and min

In summary: Okay, I'll just provide the summary for now.In summary, the problem is to find the maximum and minimum values of ##f(x, y) = 4x + y^2## subject to the constraint ##2x^2 + y^2 = 4##. There are a few different ways to approach this problem, such as solving for one variable in the constraint and then substituting it into the function, or using Lagrange multipliers. However, there appears to be some confusion with the solutions, as x = 4 and y = undefined are incorrect. It would be beneficial to sketch the ellipse and understand the domain before attempting to solve the problem.
  • #36
Pi-is-3 said:
Also, if you ever need, ## - \sqrt{a^2+b^2} \leq acos(\theta)+bsin(\theta) \leq \sqrt{a^2+b^2}##
How about this?
 
Physics news on Phys.org
  • #37
Lifeforbetter said:
How about this?

Do you mean the proof? You can prove it through Lagrange multipliers, or their is a trigonometrical proof too. For the Lagrange multipliers proof, your optimization function is ##acos(\theta)+bsin(\theta)## and restrain function is ##cos^2(\theta)+sin^2(\theta)=1##.
 

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
496
  • Calculus and Beyond Homework Help
Replies
2
Views
549
Replies
4
Views
575
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
878
  • Calculus and Beyond Homework Help
Replies
6
Views
600
  • Calculus and Beyond Homework Help
Replies
2
Views
588
  • Calculus and Beyond Homework Help
Replies
4
Views
743
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
539
Back
Top