Answer to Question #138351 in Calculus for Promise Omiponle

Question

Find the absolute maximum and absolute minimum of the function f(x,y)=2x^3+y^4 on the region {(x,y)|x^2+y^2≤4}.

If the absolute max or min occurs at multiple points list them all, separated by commas.

Absolute minimum value: ___

attained at ___.

Absolute maximum value: ___

attained at ___.

Answer 1

Absolute max and min $f(x,y)=2x^3+y^4$ on region $x^2+y^2\leq4$

first order partial derivatives:

$f_x = 6x^2, f_y=4y^3$

$f_x=0, f_y=0\implies x=0, y=0$

critical point (0,0) is inside the disk of radius 2

the function value of this critical point is $f(0,0)=0$

checking the boundaries:

$x^2+y^2=4$

$y^2=4-x^2$

$-2\leq x \leq 2$

$g(x)=2x^3+(4-x^2)^2=2x^3+16-8x^2+x^4=x^4+2x^3-8x^2+16$

$g'(x)=4x^3+6x^2-16x \implies x=\frac{-3\pm\sqrt{73}}{4}$ or 0

$x_1=0,x_2\approx-2.886, x_3\approx1.386$

the value of g(x) at the critical points and the endpoints:

$g(-2)=-16, g(2)=16, g(0)=16, g(-2.886)\approx-29.33, g(1.386)\approx9.65$

$x=-2: y^2=4-x^2=4-4=0 \implies y=0$

$x=2: y^2=4-x^2 \implies y=0$

$x=0: y^2=4-0=4 \implies y=\pm2$

$x=-2.886: y^2=4-x^2=-4.33 \implies\nexists y$

$x=1.386: y^2=4-x^2=2.08 \implies y=\pm1.44$

The function values for g(x) correspond to the following function values for f(x,y):

$g(-2)=-16 \implies f(-2,0)=-16$

$g(2)=16 \implies f(2,0)=16$

$g(0)=16 \implies f(0,2)=16, f(0,-2)=16$

$g(1.386)=9.65 \implies f(1.386,1.44)=9.65, f(1.386,-1.44)=9.65$

Comparing these values we finally have the answer:

absolute minimum value is -16 and occurs at (-2,0)

absolute maximum value is 16 and occurs at (2,0), (0,2) and (0,-2)