Max. and Min. values - I don't understand

[ChristinaMaria]

Hello!
I have been working on this problem for a while. Apparently I've found the right answer, but I don't understand why it is correct.

Homework Statement

Find the maximum and minimum value for the function
ƒ(x, y) = 2xy - 13x

on the rectangle
-3 ≤ x ≤ 3
0 ≤ y ≤ 2

The Attempt at a Solution

Critical points:

• A) ∂f/∂x = 0 = 2x - 13
• B) ∂f/∂y = 0 = 2y

• A) ∂f/∂x = 0 ⇒ x = 13/2 ⊄ [-3, 3] None within the boundary
• B) ∂f/∂y = 0 ⇒ y = 0 ∈ [0, 2], Ok!

Checking 3 ≤ x ≤ 3:
ƒ(x, 0) = -13x
ƒ(-3, 0) = 39
ƒ(3, 0) = -39

⇒ ƒ has maximum and minimum values on the rectangle at ƒ(-3, 0) = 39 and ƒ(3, 0) = -39.

--------

I checked my answer and it is correct, but I don't completely understand why. I plotted the function ƒ(x,y) =2xy - 13x and the maxima and minima A=(-3,0) and B=(3,0). The points are not even on the curve. How can the function then have maximum and minimum values at these points?

 

[PeroK]

Hello!
I have been working on this problem for a while. Apparently I've found the right answer, but I don't understand why it is correct.

Homework Statement

Find the maximum and minimum value for the function
ƒ(x, y) = 2xy - 13x

on the rectangle
-3 ≤ x ≤ 3
0 ≤ y ≤ 2

The Attempt at a Solution

Critical points:

• A) ∂f/∂x = 0 = 2x - 13
• B) ∂f/∂y = 0 = 2y

• A) ∂f/∂x = 0 ⇒ x = 13/2 ⊄ [-3, 3] None within the boundary
• B) ∂f/∂y = 0 ⇒ y = 0 ∈ [0, 2], Ok!

Checking 3 ≤ x ≤ 3:
ƒ(x, 0) = -13x
ƒ(-3, 0) = 39
ƒ(3, 0) = -39

⇒ ƒ has maximum and minimum values on the rectangle at ƒ(-3, 0) = 39 and ƒ(3, 0) = -39.

--------

I checked my answer and it is correct, but I don't completely understand why. I plotted the function ƒ(x,y) =2xy - 13x and the maxima and minima A=(-3,0) and B=(3,0). The points are not even on the curve. How can the function then have maximum and minimum values at these points?

A and B are points on the x-y plane. The function value at these points is a maximum.

You can view the function as a surface with coordinates $(x, y, f(x, y))$. Then, your maximum and minimum are points on this surface, namely:

$(-3, 0, 39)$ and $(3, 0, -39)$

 

[Orodruin]

• ∂f/∂x = 0 = 2x - 13

• B) ∂f/∂y = 0 = 2y

Double check those derivatives.

 

[ChristinaMaria]

Double check those derivatives.

Thank you

A and B are points on the x-y plane. The function value at these points is a maximum.

You can view the function as a surface with coordinates $(x, y, f(x, y))$. Then, your maximum and minimum are points on this surface, namely:
$(-3, 0, 39)$ and $(3, 0, -39)$

Thank you Sorry, I don't understand. How would you view it as a surface (x,y,f(x,y))?

 

[PeroK]

Thank you Thank you Sorry, I don't understand. How would you view it as a surface (x,y,f(x,y))?

A function of one variable $f(x)$ can be viewed as a curve above (or below) the x-axis. Similarly, a function of two variables $f(x, y)$ can be viewed as a surface about the x-y plane.

You can Google for images and videos if you want to see what this looks like, but that's what your graphing software should have done.

 

[ChristinaMaria]

A function of one variable $f(x)$ can be viewed as a curve above (or below) the x-axis. Similarly, a function of two variables $f(x, y)$ can be viewed as a surface about the x-y plane.

You can Google for images and videos if you want to see what this looks like, but that's what your graphing software should have done.

Ok, thanks again. I will look into it.

 

[Ray Vickson]

Hello!
I have been working on this problem for a while. Apparently I've found the right answer, but I don't understand why it is correct.

Homework Statement

Find the maximum and minimum value for the function
ƒ(x, y) = 2xy - 13x

on the rectangle
-3 ≤ x ≤ 3
0 ≤ y ≤ 2

The Attempt at a Solution

Critical points:

• A) ∂f/∂x = 0 = 2x - 13
• B) ∂f/∂y = 0 = 2y

• A) ∂f/∂x = 0 ⇒ x = 13/2 ⊄ [-3, 3] None within the boundary
• B) ∂f/∂y = 0 ⇒ y = 0 ∈ [0, 2], Ok!

Checking 3 ≤ x ≤ 3:
ƒ(x, 0) = -13x
ƒ(-3, 0) = 39
ƒ(3, 0) = -39

⇒ ƒ has maximum and minimum values on the rectangle at ƒ(-3, 0) = 39 and ƒ(3, 0) = -39.

--------

I checked my answer and it is correct, but I don't completely understand why. I plotted the function ƒ(x,y) =2xy - 13x and the maxima and minima A=(-3,0) and B=(3,0). The points are not even on the curve. How can the function then have maximum and minimum values at these points?
View attachment 238881

If $R$ is the rectangle $$R = \{ (x,y): -3 \leq x \leq 3, 0 \leq y \leq 2 \}$$ then a max or min over $R$ is either on the boundary of $R$ or in the interior of $R$. Any max or min in the interior of $R$ would have to be a stationary point of $f(x,y) = 2 xy - 13 x$, but as you have already shown, such a stationary point lies outside $R$. Therefore, the solution will lie on the boundary of $R$. So, all you need to is look at four problems of optimizing $f(x,y$ over the top, bottom left-side and right-side of $R$. For example, on the top the function is $f(x,2) = 2(2)x - 13 x = -9x,$ and this is to be either maximized or minimize, subject to the simple bound constraints $-3 \leq x \leq 3.$ Since we have a linear function of $x$, the maximum will occur at one end of the interval and the minimum at the other end. In other words, the two optima will occur at the top two corners of $R.$

Similarly for the optima along the bottom and the two sides of $R$. In other words, we solve the problem just by computing $f(x,y)$ at the four corners, and picking the lowest and highest of the four values.

Note: lest you think that the method is too simple-minded and needs too much work, be assured that (theoretically) some such optimization problems are difficult, and in the worst case would need evaluation at all the corners. For a problem having $10^{1000}$ corners that could involve more computing power than the entire world's resources working for the estimated age of the universe, so getting an exact solution is likely not practical.

 

[ChristinaMaria]

If $R$ is the rectangle $$R = \{ (x,y): -3 \leq x \leq 3, 0 \leq y \leq 2 \}$$ then a max or min over $R$ is either on the boundary of $R$ or in the interior of $R$. Any max or min in the interior of $R$ would have to be a stationary point of $f(x,y) = 2 xy - 13 x$, but as you have already shown, such a stationary point lies outside $R$. Therefore, the solution will lie on the boundary of $R$. So, all you need to is look at four problems of optimizing $f(x,y$ over the top, bottom left-side and right-side of $R$. For example, on the top the function is $f(x,2) = 2(2)x - 13 x = -9x,$ and this is to be either maximized or minimize, subject to the simple bound constraints $-3 \leq x \leq 3.$ Since we have a linear function of $x$, the maximum will occur at one end of the interval and the minimum at the other end. In other words, the two optima will occur at the top two corners of $R.$

Similarly for the optima along the bottom and the two sides of $R$. In other words, we solve the problem just by computing $f(x,y)$ at the four corners, and picking the lowest and highest of the four values.

Note: lest you think that the method is too simple-minded and needs too much work, be assured that (theoretically) some such optimization problems are difficult, and in the worst case would need evaluation at all the corners. For a problem having $10^{1000}$ corners that could involve more computing power than the entire world's resources working for the estimated age of the universe, so getting an exact solution is likely not practical.

Hello!
Thank you so much for your thorough explanation! I feel like I understand it a little better now. And I see now that I didn't evaluate the top left and right corners of the rectangle...

The points I found, (-3,0) and (3,0), give the maximum and minimum value of f(x,y) on the rectangle. They don't touch the graph because they are projections of the points (x,y,f(x,y,)), (-3,0,39) and (3,0,-39) on the rectangle in the xy-plane, when the figure in my graph is in three dimensions?

 

[alan2]

Good visualization is important. I much prefer this plotter. https://academo.org/demos/3d-surface-plotter/

 

[Ray Vickson]

Hello!
Thank you so much for your thorough explanation! I feel like I understand it a little better now. And I see now that I didn't evaluate the top left and right corners of the rectangle...

The points I found, (-3,0) and (3,0), give the maximum and minimum value of f(x,y) on the rectangle. They don't touch the graph because they are projections of the points (x,y,f(x,y,)), (-3,0,39) and (3,0,-39) on the rectangle in the xy-plane, when the figure in my graph is in three dimensions?

In addition to the excellent tool suggested by alan2 in #9, you can also (or instead) use the free tool Wolfram Alpha, and just enter your problem as
"plot 2*x*y-13*x over -3 <= x <= 3, 0 <= y <= 2"
(but without the " ... ")

 

[ChristinaMaria]

Thanks everyone!