Question

Local versus absolute extrema. If you recall from single-variable calculus (calculus I), if a function has only one critical point, and that critical point is a local maximum (or say local minimum), then that critical point is the global/absolute maximum (or say global/absolute minnimum). This fails spectacularly in higher dimensions (and thereís a famous example of a mistake in a mathematical physics paper because this fact was not properly appreciated.) You will compute a simple example in this problem. Let f(x; y) = e 3x + y 3 3yex . (a) Find all critical points for this function; in so doing you will see there is only one. (b) Verify this critical point is a local minimum. (c) Show this is not the absolute minimum by Önding values of f(x; y) that are lower than the value at this critical point. We suggest looking at values f(0; y) for suitably chosen y

Answer 1

Answer:

Step-by-step explanation:

Given that:

a)

$f(x,y) = e^{3x} + y^3 - 3ye^x \\ \\ \implies \dfrac{\partial f}{\partial x} = 0 = 3e^{3x} -3y e^x = 0 \\ \\ e^{2x}= y \\ \\ \\ \implies \dfrac{\partial f}{\partial y } = 0 = 3y^2 -3e^x = 0 \\ \\ y^2 = e^x$

$\text{Now; to determine the critical point:}$- $f_x = 0 ; \ \ \ \ \ f_y =0$

$\implies e^{2x} = y^4 = y \\ \\ \implies y = 0 \& y =1 \\ \\ since y \ne 0 , \ \ y = 1, \ \ x= 0\\\text{Hence, the only possible critical point= }(0,1)$

b)

$\delta = f_xx, s = f_{xy}, t = f_{yy} \\ \\ . \ \ \ \ \ \ \ \ D = rt-s^2 \\ \\ i) Suppose D >0 ,\ \ \ r> 0 \ \text{then f is minima} \\ \\ ii) Suppose \ D >0 ,\ \ \ r< 0 \ \text{then f is mixima} \\ \\ iii) \text{Suppose D} < 0 \text{, then f is a saddle point} \\ \\ iv) Suppose \ D = 0 \ \ No \ conclusion$

$Thus \ at (0,1) \\ \\ \delta = f_{xx} = ge^{3x}\implies \delta (0,1) = 6 \\ \\ S = f_{xy} = -3e^x \\ \\ \implies S_{(0,1)} = -3 \\ \\ t = f_{yy} = 6y$

$\implies t_{0,1} = 6$

$Now; D = rt - s^2 \\ \\ = (6)(6) -(-3)^2$

$= 36 - 9 \\ \\ = 27 > 0 \\ \\ r>0$

$\text{Hence, the critical point} \ (0,1) \ \text{appears to be the local minima}$

c)

$\text{Suppose we chose x = 0 and y = -3.4} \\ \\ \text{Then, we have:} \\ \\ f(0,-3.4) = 1+ (-3.4)^3 + 3(3.4) \\ \\ = -28.104 < -1$

$\text{However, if f (0,1) = 1 +1 -3 = -1 \\ \\ f(0,-3.4) = -28.104} < -1} \\ \\ \text{This explains that} -1 \text{is not an absolute minimum value of f(x,y)}$