[Math] Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ vertices

calculusmultivariable-calculus

Can't find absolute minimums and maximums!

Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ on the set D where D is the closed triangular region with vertices $(8,0), (0,6)$, and $(0,-6)$.

To find critical points:

$F_x = 0$

$F_y = 0$

$F_x$ (first order derivative with respect to x) $ = 2x – 4$

$F_y = 2y$

Setting them equal to $0$, to find critical points:

$2x – 4 = 0$

$2y = 0$

critical point: $(2,0)$

…

Now, the boundary of the triangle can be expressed in 3 lines:

Left side of triangle, $x = 0$:

Function can be expressed by the one variable function

$f(0, y) = y^2 + 9$

Absolute maximums:

You evaluate those at the given points, and I think also supposed to evaluate at the critical point? (derivative equal to $0$ etc)

$(0,6), (0,-6)$

Absolute minimums:

??? (8,0) <- why not at these points?
If y = 0, then that's when it has the absolute minimum, no?

…

Upper right side of the triangle:

between points: $(0,6)$ and $(8,0)$

$y= \frac{-3}{4} x + 6 $

Function can be expressed by the one variable function

$f(x, \frac{-3}{4} x + 6 ) = \ (\frac{-3}{4} x + 6)^2 + x^2 – 4x + 9$

Absolute maximums:

$(0, 6)$

Absolute minimums:

(8,0) INCORRECT, WHY???

…

Lower right side of the triangle:

between points: $(0,-6)$ and $(8,0)$

$y = \frac{3}{4} x – 6$

Function can be expressed by the one variable function

(3/4 x – 6)^2 + x^2 – 4x + 9

Absolute maximums:

????

Absolute minimums:

????

p.s. here's the screenshot

Best Answer

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I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.

edit 1:(add more details)

$f(x,y)=(x-2)^2+y^2+5 \implies (x-2)^2+y^2=f(x,y)-5=R^2$

you want to find max and min of $f(x,y)$ is same to find max and min of $R$.

note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$

if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.

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