Key to "Find the absolute maximum and minimum values of..."

High School
Algebra
Algebra I
Piecewise-Defined Functions

Lucille Davidson

Answered question

2021-12-08

Find the absolute maximum and minimum values of

f (x, y) = x^{2} + y^{2} - 2 x + 1

on the unit disc

x^{2} + y^{2} \leq 1

Answer & Explanation

Marcus Herman

Beginner2021-12-09Added 41 answers

Step 1
Given function is: $f (x, y) = x^{2} + y^{2} - 2 x + 1$ and disc is: $x^{2} + y^{2} \leq 1$
Step 2
Let,
$x = r \cos (0), y = r \sin (0)$
Then,
$f (r, 0) = r^{2} \cos^{2} (0) + r^{2} \sin^{2} (0) - 2 r \cos (0) + 1$
$= r^{2} - 2 r \cos (0) + 1$
Step 3
Let,
$t = \cos (0)$
Then,
$f (r, 0) = f (r, t) = r^{2} - 2 r t + 1 w i t h 0 \leq r \leq 1, - 1 \leq t \leq 1$
Taking partial derivatives,

$f_{r} = 2 r - 2 t$
$f_{t} = - 2 r$
Now, $f_{r} = 0, f_{t} = 0$ gives
r=0, t=0
Step 4
If r=0 only, then $x = r \cos (0) = 0, y = r \sin (0) = 0$
If t=0 only, then x=rt, $y = \pm r$
From this, at end point $(0, \pm 1), (1, \pm 1)$ ,
$f_{max} = 1 + 1 = 2$
$f_{min} = 1 + 1 - 2 + 1 = 1$
Step 5
Hence, the absolute maximum values is 2 and absolute minimum value is 1.

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