per the article at
economics.uwo.ca/math/resources/calculus-multivariable-functions/5-partial-derivatives-optimization-constraints/content/
dy/dx = -fx/fy = -x/(2y)
dg/dx = -gx/gy = -2xy
set the two equal and you get x=0 or y = ±1/2
checking g(x,y) at those values gives maximum f(0,1) = 2
Maximize 𝑓(𝑥, 𝑦) = 𝑥^2 + 2𝑦^2
with the constraint :
𝑦(𝑒)^𝑥^2 = 1
Do not use Langrage method
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