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the first derivative test yields complex roots.. how to proceed...
the function is
f(x,y) = (x^3) + (3*x*x*y) -15 (x*x) -15 (y*y) +72x

2007-09-10 06:17:33 · 2 answers · asked by Sudhakar A 1 in Science & Mathematics Mathematics

2 answers

f(x,y) = (x^3) + (3*x*x*y) -15 (x*x) -15 (y*y) +72x

∂f/∂x = 3x^2 + 6 xy - 30x + 72 = 0
∂f/∂y = 3x^2 - 30y = 0

since y = 3x^2/30 = x^2/10

3x^2 + 3x^3/5 - 30 x + 72 = 0 multiply by 5/3 to get
x^3 + 5x^2 - 50x + 120 = 0

I believe you are correct that this has no real roots. In that case, the function is monotonic and has no extreme points.

2007-09-10 06:35:08 · answer #1 · answered by anobium625 6 · 1 0

I did not do the calculations but a 3rd degree polynom with real coefficients has ALWAYS at least one real root.
df/dx= 3x^2 +6xy -30x+72=0
df/dy= 3x^2-30y=0 so y =x^2/10
3x^2+3x^3/5 -30x+72 =0 is really a 3rd degree equation

2007-09-10 07:17:33 · answer #2 · answered by santmann2002 7 · 0 0

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