Christopher A. answered • 12/10/19

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Your work so far is correct. Unfortunately, the second derivative test is inconclusive when D = 0. In this case, we can show that there is a global maximum for the function.

First, we observe that x^{4} ≥ 0 for all x and y^{4} ≥ 0 for all y. Therefore, (x^{4} + y^{4}) ≥ 0 for all x,y.

Next we manipulate our function to show that it has a maximum:

f(x,y) = 1 - x^{4}- y^{4}

= 1 - (x^{4}+ y^{4}) (factor -1 out of the 2nd and 3rd terms)

f(x, y) + (x^{4}+ y^{4}) = 1 (add (x^{4}+ y^{4}) to both sides)

(x^{4}+ y^{4}) = 1 - f(x,y) (subtract f(x,y) from both sides)

1 - f(x,y) ≥ 0 (substitute for (x^{4}+ y^{4}) in the inequality (x^{4}+ y^{4}) ≥ 0)

1 ≥ f(x,y) (add f(x,y) to both sides)

f(x,y) ≤ 1

Now that we have shown a global maximum, we can show that (0,0) is a relative maximum.

f(0,0) = 1

f(x,y) ≤ f(0,0)

Therefore, (0,0) is a relative maximum.