Lesson Finding the minimum of a function defined on a curve in a coordinate plane

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Finding the minimum of a function defined on a curve in a coordinate plane


Problem 1

The positive variables  x  and  y  are such that  x%5E4y = 32.
A function  z  is defined by   z = x%5E2+%2B+y.
Find the values of  x  and  y  that give  z  a stationary value and show that this value of  z  is a minimum.

Solution

x%5E4y+=+32 --> y+=+32x%5E%28-4%29

z+=+x%5E2+%2B+y = x%5E2+%2B+32x%5E%28-4%29       (1)

dz%2Fdx = 2x+-+128x%5E%28-5%29            (2)

d%5E2z%2Fdx%5E2 = 2+%2B+640x%5E%28-6%29            (3)


The stationary point is where the derivative is zero.

2x-128x%5E%28-5%29 = 0

2x = 128x%5E%28-5%29

x%5E6 = 128%2F2+=+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%2Fx%5E4 = 32%2F16 = 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.


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


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



    

            Plot  z = x%5E2 + 32x%5E%28-4%29  (see formula (1)


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