Finding minimum for an equation with two variables

In summary: Then I turned x^2 + 4*x*(y - 1) into a square: [(x + 2*(y - 1)]^2 and subtracted [2*(y - 1)]^2, which is 4*(y - 1)^2 to balance it out, so I got:[x + 2*(y - 1)]^2 - 4*(y - 1)^2 + 5*y^2 - 6*y + 7However when I complete the square for the other part I get:[x +
  • #1
Hivoyer
27
0

Homework Statement



I have the equation: x^2 + 2*x*y + 5*y^2 - 4*x - 6*y +7 and I have to find the minimum value
I'm getting something that looks half like the correct answer, but not quite right...

Homework Equations



The answer from the answer book is:

[x + 2*(y - 1)]^2 + (y + 1)^2 + 2

The Attempt at a Solution



Ok first I took 2*x*y and -4*x and turned them into 4*x*(y - 1), so I got:

x^2 + 4*x*(y - 1) + 5*y^2 - 6*y + 7

Then I turned x^2 + 4*x*(y - 1) into a square: [(x + 2*(y - 1)]^2 and subtracted [2*(y - 1)]^2, which is 4*(y - 1)^2 to balance it out, so I got:

[x + 2*(y - 1)]^2 - 4*(y - 1)^2 + 5*y^2 - 6*y + 7

However when I complete the square for the other part I get:

[x + 2*(y - 1)]^2 - 4*(y - 1)^2 + 5*[(y - 3)^2 - 9] + 7

when then gives me:

[x + 2*(y - 1)]^2 - 4*(y - 1)^2 + 5*(y - 3)^2 - 38

and this is not what the answer in the answer book I've written above is.Where did I go wrong?
 
Physics news on Phys.org
  • #2
Hivoyer said:

Homework Statement



I have the equation: x^2 + 2*x*y + 5*y^2 - 4*x - 6*y +7 and I have to find the minimum value
I'm getting something that looks half like the correct answer, but not quite right...

Homework Equations



The answer from the answer book is:

[x + 2*(y - 1)]^2 + (y + 1)^2 + 2

That doesn't look like a "minimum value" of anything. Please give the exact wording of the problem from your text.
 
  • #3
LCKurtz said:
That doesn't look like a "minimum value" of anything. Please give the exact wording of the problem from your text.

Here is the problem from the book(part (c) in the red rectangle):
a3bd5adaf8a79c6b.png


Here is the solution of (c) from the answer book(again surrounded in red):
cd25dbf3ab519614.png
 
  • #4
Hivoyer said:

Homework Statement



I have the equation: x^2 + 2*x*y + 5*y^2 - 4*x - 6*y +7 and I have to find the minimum value
I'm getting something that looks half like the correct answer, but not quite right...

Homework Equations



The answer from the answer book is:

[x + 2*(y - 1)]^2 + (y + 1)^2 + 2

The Attempt at a Solution



Ok first I took 2*x*y and -4*x and turned them into 4*x*(y - 1),

##2xy-4x\ne 4x(y-1)##
 

Related to Finding minimum for an equation with two variables

1) How do I find the minimum value of an equation with two variables?

To find the minimum value of an equation with two variables, you can use the method of partial derivatives. This involves taking the partial derivative of the equation with respect to each variable and setting them equal to zero. The resulting system of equations can then be solved to find the values of the variables that minimize the equation.

2) Can I use graphical methods to find the minimum of an equation with two variables?

Yes, graphical methods can also be used to find the minimum of an equation with two variables. This involves graphing the equation and identifying the lowest point on the graph, which corresponds to the minimum value. However, this method may not always be accurate and it is recommended to use the method of partial derivatives for more precise results.

3) What is the significance of finding the minimum value of an equation with two variables?

Finding the minimum value of an equation with two variables is important in many real-world applications, such as optimization problems in economics and engineering. It allows us to determine the lowest possible value of a given function, which can help in making informed decisions and improving efficiency.

4) Are there any limitations to finding the minimum value of an equation with two variables?

Yes, there are some limitations to finding the minimum value of an equation with two variables. In some cases, there may not be a minimum value or the minimum value may not be unique. This can happen when the equation has multiple local minima or when the minimum value is at the boundary of the feasible region.

5) Can I use the method of partial derivatives to find the minimum of any equation with two variables?

Yes, the method of partial derivatives can be used to find the minimum of any equation with two variables, as long as the equation is differentiable. This means that the equation must have a continuous first derivative with respect to both variables. If this condition is not met, the method may not be applicable.

Similar threads

Solve the simultaneous equations involving x, y, and their squares
    • Precalculus Mathematics Homework Help
Replies
2
Views
614
Obtain the digits ## x ## and ## y ##
    • Precalculus Mathematics Homework Help
Replies
3
Views
860
Find the range of the function ##f(x) = \frac{e^{2x}-e^x+1} {e^{2x}+e^x+1}##
    • Precalculus Mathematics Homework Help
Replies
8
Views
342
Solve ##\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{x}}=4\dfrac{1}{4} \cdots##
    • Precalculus Mathematics Homework Help
Replies
10
Views
696
Solve the given simultaneous equations in x^2, x and y
    • Precalculus Mathematics Homework Help
Replies
1
Views
627
Express ##x^3+y^3## in terms of ##m## and ##n##
    • Precalculus Mathematics Homework Help
Replies
17
Views
533
Finding a Mistake: Probability of Y with X and Spins
    • Precalculus Mathematics Homework Help
Replies
7
Views
2K
Obtain the eight incongruent solutions of the linear congruence
    • Precalculus Mathematics Homework Help
Replies
2
Views
1K
To find the boundedness of a given function
    • Precalculus Mathematics Homework Help
Replies
13
Views
700
Solve the simultaneous equations
    • Precalculus Mathematics Homework Help
Replies
7
Views
1K

Hot Threads

Recent Insights

Back
Top