# Thread: Critical values of Z=(x-2)^4+(y-3)^4.

1. ## Critical values of Z=(x-2)^4+(y-3)^4.

Z=(x-2)^4+(y-3)^4
plz help me 2 find out the critical and extreme values

2. Originally Posted by powerxxx
Z=(x-2)^4+(y-3)^4
plz help me 2 find out the critical and extreme values
Solve simultaneously:

$\frac{\partial z}{\partial x} = 0$

$\frac{\partial z}{\partial y} = 0$

If you need more help, please show all your work and say exactly where you get stuck.

3. I’m new... could you plz slow me the step by step process for this problem

4. Originally Posted by powerxxx
I’m new... could you plz slow me the step by step process for this problem
If you do not know how to at least get the partial derivatives, then I suggest you forget about this question for a while and go back and review what you have already been taught about partial differentiation (the question assumes that you know it).

Otherwise, please show all your working so far and say where you are stuck.

5. Originally Posted by powerxxx
I’m new... could you plz slow me the step by step process for this problem
We understand that you are new but you must show us an attempt at the problem to prove you are trying.

Find

$f_x$
$f_{xx}$
$f_y$
$f_{yy}$
$f_{xy}$

Then use

$(f_{xx})(f_{yy}) -(f{xy})^2$

6. f_X=4(X-2)^3
f_XX=12(X-2)^2
f_XY=0

f_Y=4(Y-3)^3
f_YY=12(Y-3)^2
F_YX=0

but how i will get the extreme value

7. Originally Posted by powerxxx
f_X=4(X-2)^3
f_XX=12(X-2)^2
f_XY=0

f_Y=4(Y-3)^3
f_YY=12(Y-3)^2
F_YX=0

but how i will get the extreme value
Ok now set your 1st partial derivatives to 0

$4(x-2)^3 = 0$

x = 2

$4(y-3)^3 = 0$

y = 3

so your critical value is (2,3)

Now use

$(f_{xx})(f_{yy}) -(f{xy})^2 = D$

if D = 0 inconclusive
if D > 0 max or min
if D < 0 saddle point