Another Absolute Maximum Question

[jegues]

Homework Statement

Find the maximum value of the function $$f(x, y) = x^{2} - y^{2} + x$$ considering only points inside and on the boundary of the region bounded by the curve,

$$x = \sqrt{1-y^{2}}, x=0$$

Homework Equations

The Attempt at a Solution

See figure attached for my attempt.

I drew a quick sketch of the region given and I've found it to be a circle with a vertical line passing through its center. That being said, I didn't know what "half" of the circle was my enclosed region. Is it the left half or the right half? How do you distinguish this?

I continued working while only considering the "right half" as my region.

I found the critical points inside my region and continued to look for critical points along the edges of my region.

I labeled these accordingly, C1 and C2.

I still haven't obtained the correct answer of 2, so I'm not entirely sure what I've done wrong.

Could someone help me out?

 

[tiny-tim]

hi jegues!

… I didn't know what "half" of the circle was my enclosed region. Is it the left half or the right half?

you haven't read the question carefully enough …

√ means the positive square root (or zero), so it has to be the right-hand half!

I found the critical points inside my region and continued to look for critical points along the edges of my region.

you were ok up to h' = 4x + 1 …

since that's always positive (in the region), h reaches a maximum at the maximum boundary of x

 

[jegues]

hi jegues!


you haven't read the question carefully enough …

√ means the positive square root (or zero), so it has to be the right-hand half!


you were ok up to h' = 4x + 1 …

since that's always positive (in the region), h reaches a maximum at the maximum boundary of x

I'm not entirely sure what you mean by maximum boundary of x.

Do you mean the largest value of x I can obtain in the region provided? That would simply be x=1.

Therefore h(1) = 2, which would correspond to my absolute max, correct?

 

[tiny-tim]

Yup!

 

[jegues]

Why is this different then the standard procedure when checking for critical points along the edge of a region?

I'm used to having to take a derivative and equate it to 0 in order to figure out my point of interest.

What makes this different?

 

[tiny-tim]

hi jegues!

(just got up :zzz: …)

Why is this different then the standard procedure when checking for critical points along the edge of a region?

I'm used to having to take a derivative and equate it to 0 in order to figure out my point of interest.

What makes this different?

It isn't different …

you equate the derivative to zero along the semi-circle boundary …

(if there's a solution, then of course that's your solution, but if there isn't a zero, then …)

with no zero, the solution must be at the boundary of the semi-circle boundary …

looking at it, you'd say the boundary is the two points on the y-axis, but that's wrong because the derivative was wrt x, and x goes from 0 to 1, so the boundary is at x = 0 or 1

(and since the derivative wrt x is positive, the maximum must be at x = 1)