Question 1193060
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            The solution by @greenestamps needs to be corrected.


            I came to bring a correct solution.



<pre>

{{{x^4y = 32}}} --> {{{y = 32x^(-4)}}}

{{{z = x^2 + y}}} = {{{x^2 + 32x^(-4)}}}       (1)

{{{dz/dx}}} = {{{2x - 128x^(-5)}}}            (2)

{{{d^2z/dx^2}}} = {{{2 + 640x^(-6)}}}            (3)


The stationary point is where the derivative is zero.

{{{2x-128x^(-5)}}} = {{{0}}}

{{{2x}}} = {{{128x^(-5)}}}

{{{x^6}}} = {{{128/2 = 64}}}

x = 2    (actually, x = +/- 2, but since we consider everything in positive numbers, we take x = 2).

At the stationary point, {{{x = 2}}} and {{{y}}} = {{{32/x^4}}} = {{{32/16}}} = {{{2}}}


    The stationary point is a minimum if the second derivative at the point is positive; 
    or it is a maximum if that derivative is negative.  
    At x = 2, the second derivative is OBVIOULSLY positive (it is clear without any calculations)


So the stationary point is a minimum.


ANSWER: z has a stationary point that is a minimum when x = 2 and y = 2.


To make this result visually verifiable, I prepared a plot below.



    {{{graph( 400, 400, -5, 5,  -5, 20,
              x^2 + 32x^(-4)
)}}}

            Plot  z = {{{x^2}}} + {{{32x^(-4)}}}  (see formula (1)
</pre>

Solved.