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 May 23rd, 2017, 12:51 PM #1 Member   Joined: Jan 2016 From: Blackpool Posts: 95 Thanks: 2 Auxiliary equation What is the minimum value of yx^3 on the circle x^2 + y^2 = 4? I started by using the auxiliary equation to get two expressions for the partial derivatives of x and y and got for y: 0 = (x^3) - 2λy x: 0 = (3x^2) - 2λy I am having trouble now cancelling out either the x or y term and was wondering if someone could help, thanks! Last edited by skipjack; May 23rd, 2017 at 04:16 PM.
 May 23rd, 2017, 01:27 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,276 Thanks: 2437 Math Focus: Mainly analysis and algebra I would parametrise the circle in terms of a single variable $t$ and then find the extrema of $y(t)x^3(t)$, a function of $t$ only.

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