Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ vertices
$begingroup$
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
calculus multivariable-calculus
$endgroup$
add a comment |
$begingroup$
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
calculus multivariable-calculus
$endgroup$
add a comment |
$begingroup$
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
calculus multivariable-calculus
$endgroup$
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
calculus multivariable-calculus
calculus multivariable-calculus
asked Mar 2 '16 at 0:28
JackJack
311615
311615
add a comment |
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1 Answer
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$begingroup$
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.
$endgroup$
$begingroup$
what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
$endgroup$
– Jack
Mar 2 '16 at 3:22
add a comment |
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
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.
$endgroup$
$begingroup$
what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
$endgroup$
– Jack
Mar 2 '16 at 3:22
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
$endgroup$
– Jack
Mar 2 '16 at 3:22
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Mar 2 '16 at 6:15
answered Mar 2 '16 at 0:57
chenbaichenbai
6,60211024
6,60211024
$begingroup$
what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
$endgroup$
– Jack
Mar 2 '16 at 3:22
add a comment |
$begingroup$
what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
$endgroup$
– Jack
Mar 2 '16 at 3:22
$begingroup$
what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
$endgroup$
– Jack
Mar 2 '16 at 3:22
$begingroup$
what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
$endgroup$
– Jack
Mar 2 '16 at 3:22
add a comment |
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