SOLUTION: The positive variables x and y are such that x^4 y = 32. A third variable z id defined by z = x^2 + y. Find the values x and y that give z a stationary value and show that this va

Question 1194170: The positive variables x and y are such that x^4 y = 32. A third variable z id defined by z = x^2 + y.
Find the values x and y that give z a stationary value and show that this value of z is a minimum.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
x^4 y = 32 -> y = 32/x^4
Thus z = x^2 + y -> z = x^2 + 32/x^4
Stationary values are the points where dz/dx = 0:
dz/dx = 0 = 2x - 128/x^5
To solve for x, multiply through by x^5:
2x^6 - 128 = 0 -> x^6 = 64
Solutions: x = 2, -2
Thus y = 32/(2)^4 = 2
The stationary points are (2,2) and (-2,2)
Computing the 2nd derivative, since d^2z/dx^2 = 2 + 640/x^6 is positive at x=2, -2, the points are minima

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