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

May 2010
3
0
Z=(x-2)^4+(y-3)^4
plz help me 2 find out the critical and extreme values
 
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mr fantastic

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Dec 2007
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Z=(x-2)^4+(y-3)^4
plz help me 2 find out the critical and extreme values
Solve simultaneously:

\(\displaystyle \frac{\partial z}{\partial x} = 0\)

\(\displaystyle \frac{\partial z}{\partial y} = 0\)

If you need more help, please show all your work and say exactly where you get stuck.
 
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May 2010
3
0
I’m new... could you plz slow me the step by step process for this problem
 

mr fantastic

MHF Hall of Fame
Dec 2007
16,948
6,768
Zeitgeist
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.
 
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Jul 2007
894
298
New Orleans
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

\(\displaystyle f_x\)
\(\displaystyle f_{xx}\)
\(\displaystyle f_y\)
\(\displaystyle f_{yy}\)
\(\displaystyle f_{xy}\)

Then use

\(\displaystyle (f_{xx})(f_{yy}) -(f{xy})^2\)
 
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May 2010
3
0
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
 
Jul 2007
894
298
New Orleans
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

\(\displaystyle 4(x-2)^3 = 0\)

x = 2

\(\displaystyle 4(y-3)^3 = 0\)

y = 3

so your critical value is (2,3)

Now use

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

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