Sophie G.

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

Find the absolute extrema for f(x,y)=4-x^2-y^4+1/2y^2 over the closed disk D:x^2+y^2 is less than or equal to 1

2 Answers By Expert Tutors

By:

Kevin S. answered • 08/15/20

Tutor
4.9 (528)

Calculus Instructor with PhD

Yefim S. answered • 08/13/20

Tutor
5 (20)

Math Tutor with Experience

Tom K.

tutor
Yefim has a great answer as always. Note that, as all variables are squared throughout, we could let u = x^2 and v = y^2 and consider f(u,v) = 4 - u - v^2 + 1/2 v on u + v &lt;= 1, u, v &gt;= 0. Given that df/du = -1, we note that all maxima will be on u = 0 and all minima on u + v = 1. For u = 0, we substitute and take the derivative with respect to v and thus find the maximum at v = 1/4 (note that the second derivative is negative). On u + v = 1, we can substitute for one variable in terms of the other. Remember to check the endpoints as well.
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08/14/20

Kevin S.

tutor
Nice and clean, but why can't y=0, when x can? The minimum should be f(1,0)=f(-1,0)=3, no?
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08/15/20

Yefim S.

Sorry I lost case y = 0 and x = 1 and x = - 1 and absolute min = 3
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08/15/20

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