# Finding minimum value of function with two variable

• MHB
• gevni
In summary: Yes, x can be anywhere between 0 and the length of the side of the square, and y can be anywhere between 0 and the length of the side of the square.
gevni
I have a formula for cost calculation that contain x and y two variable. I have to find the value of (x,y) where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value.
I took 1st partial derivative with respect to x and then with y and found the value of x and y where function will give minimum value. My question is how to combine both to get one answer as optimal (x,y) where function has minimum cost?

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gevni said:
I have a formula for cost calculation that contain x and y two variable. I have to find the value of (x,y) where that formula will gives minimum value as cost should not be equal to zero, it has some minimum value.
I took 1st partial derivative with respect to x and then with y and found the value of x and y where function will give minimum value. My question is how to combine both to get one answer as optimal (x,y) where function has minimum cost?
I assume you set the partial derivatives to 0 and solved the two resulting equations for x and y. Now what do mean by "combine both to get one answer? The one point, (x, y) IS the point where f(x, y) is minimum (or maximum- you need to check which). For example if $f(x,y)= (x-3)^2+(y- 4)^2$ then setting the partial derivatives equal to 0 gives $2(x- 3)= 0$ and $2(y- 4)= 0$ so x= 3 and y= 4. The "critical point" is (3, 4) where the value of the function is f(3, 4)= 0, a minimum.

Country Boy said:
I assume you set the partial derivatives to 0 and solved the two resulting equations for x and y. Now what do mean by "combine both to get one answer? The one point, (x, y) IS the point where f(x, y) is minimum (or maximum- you need to check which). For example if $f(x,y)= (x-3)^2+(y- 4)^2$ then setting the partial derivatives equal to 0 gives $2(x- 3)= 0$ and $2(y- 4)= 0$ so x= 3 and y= 4. The "critical point" is (3, 4) where the value of the function is f(3, 4)= 0, a minimum.
Thank you! yes after getting partial derivatives with x and y I set the resulting equations to 0 for getting values of x and y and got the formula for x= ...and y=...
Just want to confirm that now I have to replace x and y in original cost equation with these formulas (x= and y=) and it will results in minimum cost for that cost equation right?.
One more question how do I know that these values of x and y are for minimum?

One way is to compare the value at that point to values close by. Another is to use the "second derivative test": if $(f_{xx})(f_{yy})-(f_{xy})^2$ is positive at the given point, then it is either a maximum or a minimum. If it is negative then the point is a "saddle point".

(This is the determinant of the "Hessian matrix", $\begin{bmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy}\end{bmatrix}$. It is actually the product of the two eigenvalues. Each eigenvalue tells you if the graph curves up (if the eigenvalue is positive) or down (if negative) in the x or y directions. The determinant is positive if they have the same sign, both positive or both negative, negative if they have different signs.)

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Country Boy said:
One way is to compare the value at that point to values close by. Another is to use the "second derivative test": if $(f_{xx})(f_{yy})-(f_{xy})^2$ is positive at the given point, then it is either a maximum or a minimum. If it is negative then the point is a "saddle point".

Thank you I compared points closed by and get to know the minimum. Now there is one more thing, I have to provide the range of x and y within which if their values fall then there is a solution else there will be no solution. Can you please look into the attached figure it is n*n square divided into A,B and C area, can you please help me finding the upper range for both variables (x,y)?

0<x< ?
0<y<?

#### Attachments

• 110461264_1177808379226535_2654335419636091325_n.jpg
38.5 KB · Views: 49
That picture is very hard to understand. You have region labeled "A", "B", "C", and "A" and are apparently using those letters to represent the areas of those regions but don't say that. In addition you have an "h" (or is it "n"?) but do not define it. Is h the length of a side of the square? Finally, what in the world does this picture have to do with the original problem, a "cost calculation"?

Country Boy said:
That picture is very hard to understand.
I think I batter had to ask it as a separate question (with detail) as it is a different problem. Many thanks for your help :)

## 1. How do I find the minimum value of a function with two variables?

To find the minimum value of a function with two variables, you can use a method called gradient descent. This involves taking the partial derivatives of the function with respect to each variable and using them to iteratively update the values of the variables until the minimum value is reached.

## 2. Can I use calculus to find the minimum value of a function with two variables?

Yes, you can use calculus to find the minimum value of a function with two variables. This involves taking the partial derivatives of the function and setting them equal to zero to find the critical points. Then, you can use the second derivative test to determine if these critical points are minimum values.

## 3. What is the difference between local and global minimum values of a function with two variables?

A local minimum value is the smallest value of a function within a specific range of values for the variables. A global minimum value is the smallest value of the function across all possible values for the variables. In other words, a global minimum value is also a local minimum value, but a local minimum value may not necessarily be a global minimum value.

## 4. Can I use a computer program to find the minimum value of a function with two variables?

Yes, there are many computer programs and software packages that can help you find the minimum value of a function with two variables. These programs use various algorithms and methods to calculate the minimum value, such as gradient descent, Newton's method, or genetic algorithms.

## 5. Is the minimum value of a function with two variables always unique?

No, the minimum value of a function with two variables may not always be unique. In some cases, there may be multiple local minimum values or no minimum value at all. It is important to carefully analyze the function and its critical points to determine the uniqueness of the minimum value.

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