What is the extreme y(t,β)=-4e^{-4t}(β+15)+5e^{-5t}(β+12) ?

The extreme y(t,β)=-4e^{-4t}(β+15)+5e^{-5t}(β+12) is Saddle(ln(5/4),0)

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extreme y(t,β)=-4e^{-4t}(β+15)+5e^{-5t}(β+12)

extreme points

y (t, β) = - 4 e^{- 4 t} (β + 15) + 5 e^{- 5 t} (β + 12)

Saddle (ln (\frac{5}{4}), 0)

Solution steps

Find the critical points:

(ln (\frac{5}{4}), 0)

\frac{\partial ^{2} f}{\partial t ^{2}} = 0

D (t, β) = - e^{- 10 t} (16 e^{t} - 25)^{2}

Check the sign of at each critical point

Check critical point

(ln (\frac{5}{4}), 0) :

Saddle

Saddle (ln (\frac{5}{4}), 0)

f (x, y) = 2 x^{2} + 2 y^{2} - 4 y + 9

f (x) = - x^{2} - 2 x - 3

f (x) = x (x^{3} - x + 10)

f (x) = - 462 x + 64 x^{2} - 2 x^{3}

f (x, y) = 3 x^{3} y - 2 x^{2} y^{2} + 3 y

What is the extreme y(t,β)=-4e^{-4t}(β+15)+5e^{-5t}(β+12) ?
The extreme y(t,β)=-4e^{-4t}(β+15)+5e^{-5t}(β+12) is Saddle(ln(5/4),0)