Calculus

[keylah]

Hi! So this question is on my midterm review and I'm not quite sure how to solve it.

Find the points on the surface xy3
+z2=4 that are closest to the origin.

Here's what I have so far:
Origin means (0,0,0)
So: D=√x2+y2+z2
D2= x2+y2+z2
and
z2=4-xy3
f(x,y)= x2+y2+4-xy3

fx= 2x-y3
fy= 2y-3xy2
fxx= 2
fyy= 2-6xy
fxy= -3y2Working with these two first,

Comments

[derralf]

you don't need Lagrange multipliers. the function f(x,y) is the square-distance to the origin and x and y are not restricted.

1) tripple check every step you make. just one mistake when you transcribe a formula and the whole exercise is screwed up.

2) everytime you divide both sides of an equation you do this under the assumption that the divisor is not equal to zero. The case that it is zero has to be treated seperately.

3) y^4=c with c>0 has two real valued solutions : y=c^(1/4) and y=-c^(1/4).

4) you have already found that x=1/2 y^3. So every solution for y gives you a corresponding value for x.

5) check f(x,y) for each critical point and select the point with the smallest square-distance as the answer. You don't need to check the Hessian matrix if you don't want to classify the critical points.

alternatively : instead of doing the full multivariate optimization, you could directly plug your result x=1/2 y^3 into the formula for the distance and minimize

f(y) = 4 + y^2 - 1/4 y^6