Solve Optimization Problem: Find Min & Max of f(x,y)

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SUMMARY

The optimization problem involves finding the minimum and maximum of the function \(f(x,y)=4x^{2}-y^{2}-xy-2x+6y\) under the constraints \(y=4-2x\), \(x\geq 0\), and \(y\geq -2\). The minimum value identified is \(f_{\min}=\frac{15}{2}\) and the maximum value is \(f_{\max}=20\) using Lagrange multipliers. The initial minimum found at \(f(0,-2)=-16\) is invalid as it lies outside the defined region. The discrepancy with the provided answer sheet indicates potential errors in the published solutions.

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Yankel
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Hello all

I am trying to find minimum and maximum of the following function:

\[f(x,y)=4x^{2}-y^{2}-xy-2x+6y\]

under the constraints:

\[y=4-2x\]

\[x\geq 0\]

\[y\geq -2\]I tried solving this problem using the method of the method of bounded and closed domain, understanding that the constraints creates a triangle. I checked every line in the triangle, the edge points and the local min and max for each line (if there were any).

I got that the minimum value was f(0,-2)=-16 and the maximum was f(3,-2)=20
(should I have used Lagrange multipliers ?)

The problem is:

1. I entered this problem to MAPLE, and got max like mine, but min at f(0.5,3)=7.5. I found this point, but it isn't the absolute minimum.

2. In the answers sheet for this problem there are 4 possible answers for the sum of the min+max: 30.5, -7, 0. 16.
Non of them are according to my solution or MAPLE's.

Can you please assist me with solving this problem ?

Thank you !
 
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Using Lagrange multipliers, I find:

$$f_{\min}=\frac{15}{2}$$

$$f_{\max}=20$$

Post your work using Lagrange, and I will be glad to look it over. :D
 
I didn't use Lagrange, I looked at it as an optimization problem in a closed region, using a triangle.

I notice that the point (0,-2) which is my smallest, is not even in the region the constraint creates. Therefore, if I ignore it, I get the same results you got using Lagrange. This means that the published answers are wrong (if you, me and MAPLE say the same thing...).
 
Hi,

It's even easier if you simply use the equality constrain to go through a single valued function and derive the function.

I agree with your answers, so maybe the provided ones are wrong.
 

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