MATH SOLVE

2 months ago

Q:
# We want to find the absolute maximum value of f(x, y) = x 3 − 3x − y 2 + 12 on the domain d = {(x, y) | x 2 + y 2 ≤ 9, y ≥ 0}, a half disk. (a) (3 pts) first, find the critical points of f, and determine if any of the critical points is a local maximum in the domaind. (b) (3 pts) then, find the (absolute) maximum value of f along the boundary ofd.the bottom edge of d and along the top arc ofd. (note: some of the values might be a bit messy. give solutions to two decimal places.) (c) (2 pts) using parts (a) and (b), find the absolute maximum value of f on d

Accepted Solution

A:

[tex]f(x,y)=x^3-3x-y^2+12[/tex]

Critical points occur where both first order derivatives vanish:

[tex]f_x=3x^2-3=0\implies x=\pm1[/tex]

[tex]f_y=-2y=0\implies y=0[/tex]

So we have two critical points, (-1, 0) and (1, 0), both of which lie in the domain (on the boundary [tex]y=0[/tex]). At these points, we get values of [tex]f(-1,0)=14[/tex] and [tex]f(1,0)=10[/tex].

If we focus on just the boundaries:

(1) If [tex]y=0[/tex], then [tex]f(x,0)=g(x)=x^3-3x+12[/tex], and [tex]g'(x)=3x^2-3=0\implies x=\pm1[/tex], so there are no other critical points along this boundary.

(2) If we take [tex]x=3\cos t[/tex] and [tex]y=3\sin t[/tex], we can parameterize the semicircular arc by considering [tex]0\le t\le\pi[/tex]. So we can write

[tex]f(x,y)=f(x(t),y(t))=h(t)=27\cos^3t-9\cos t-9\sin t+12[/tex]

[tex]h'(t)=-9\sin t(9\cos^2t+2\cos t-1)=0[/tex]

[tex]\implies\begin{cases}\sin t=0\\9\cos^2t+2\cos t-1=0\end{cases}[/tex]

In the first case, we have either [tex]t=0[/tex] or [tex]t=\pi[/tex]. In the second, we can solve the quadratic to find [tex]t\approx1.33[/tex] or [tex]t\approx2.05[/tex]. At each critical point, we get

[tex]h(0)=30[/tex]

[tex]h(\pi)=-6[/tex]

[tex]h(1.33)\approx1.73[/tex]

[tex]h(2.05)\approx6.42[/tex]

Over the domain [tex]\mathcal D[/tex], we get an absolute maximum value of 30 at [tex]t=0[/tex], which corresponds to the point [tex](x,y)=(3,0)[/tex].

Critical points occur where both first order derivatives vanish:

[tex]f_x=3x^2-3=0\implies x=\pm1[/tex]

[tex]f_y=-2y=0\implies y=0[/tex]

So we have two critical points, (-1, 0) and (1, 0), both of which lie in the domain (on the boundary [tex]y=0[/tex]). At these points, we get values of [tex]f(-1,0)=14[/tex] and [tex]f(1,0)=10[/tex].

If we focus on just the boundaries:

(1) If [tex]y=0[/tex], then [tex]f(x,0)=g(x)=x^3-3x+12[/tex], and [tex]g'(x)=3x^2-3=0\implies x=\pm1[/tex], so there are no other critical points along this boundary.

(2) If we take [tex]x=3\cos t[/tex] and [tex]y=3\sin t[/tex], we can parameterize the semicircular arc by considering [tex]0\le t\le\pi[/tex]. So we can write

[tex]f(x,y)=f(x(t),y(t))=h(t)=27\cos^3t-9\cos t-9\sin t+12[/tex]

[tex]h'(t)=-9\sin t(9\cos^2t+2\cos t-1)=0[/tex]

[tex]\implies\begin{cases}\sin t=0\\9\cos^2t+2\cos t-1=0\end{cases}[/tex]

In the first case, we have either [tex]t=0[/tex] or [tex]t=\pi[/tex]. In the second, we can solve the quadratic to find [tex]t\approx1.33[/tex] or [tex]t\approx2.05[/tex]. At each critical point, we get

[tex]h(0)=30[/tex]

[tex]h(\pi)=-6[/tex]

[tex]h(1.33)\approx1.73[/tex]

[tex]h(2.05)\approx6.42[/tex]

Over the domain [tex]\mathcal D[/tex], we get an absolute maximum value of 30 at [tex]t=0[/tex], which corresponds to the point [tex](x,y)=(3,0)[/tex].