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Molly's question at Yahoo! Questions regarding constrained optimization

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MarkFL

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Feb 24, 2012
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Here is the question:

Constrained optimization and Lagrange Multipliers? Help please!?


Hi! Please help me with the question below!

Find the minimum value of the function f(x,y)=12x^2+7y^2+6xy+8x+2y+4 subject to the constraint 4x^2+2xy=1

Minimum Value: _____

THANK YOU!! :)
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello Molly,

We are given the objective function:

\(\displaystyle f(x,y)=12x^2+7y^2+6xy+8x+2y+4\)

subject to the constraint:

\(\displaystyle g(x,y)=4x^2+2xy-1=0\)

Now, using Lagrange multipliers, we obtain:

\(\displaystyle 24x+6y+8=\lambda\left(8x+2y \right)\)

\(\displaystyle 14y+6x+2=\lambda\left(2x \right)\)

This implies:

\(\displaystyle \lambda=\frac{12x+3y+4}{4x+y}=\frac{7y+3x+1}{x}\)

Cross-multiplication yields:

\(\displaystyle 12x^2+3xy+4x=28xy+12x^2+4x+7y^2+3xy+y\)

This reduces to:

\(\displaystyle 0=28xy+7y^2+y=y(28x+7y+1)\)

This gives us two cases to consider:

i) \(\displaystyle y=0\)

Substituting into the constraint, we find:

\(\displaystyle 4x^2-1=0\implies x=\pm\frac{1}{2}\)

Hence, we find the two critical points:

\(\displaystyle \left(0,\pm\frac{1}{2} \right)\)

ii) \(\displaystyle 28x+7y+1=0\implies y=-\frac{28x+1}{7}\)

Substituting into the constraint, we find:

\(\displaystyle 4x^2+2x\left(-\frac{28x+1}{7} \right)-1=0\)

Multiply through by 7:

\(\displaystyle 28x^2-56x^2-2x-7=0\)

\(\displaystyle 28x^2+2x+7=0\)

Observing that the discriminant is negative, we know there are no real roots, and so this case yields no critical points.

Now, we check the value of the objective function at the two critical points found in the first case:

\(\displaystyle f\left(-\frac{1}{2},0 \right)=3+0+0-4+0+4=3\)

\(\displaystyle f\left(\frac{1}{2},0 \right)=3+0+0+4+0+4=11\)

And so we may conclude that:

\(\displaystyle f_{\min}=f\left(-\frac{1}{2},0 \right)=3\)

We cannot conclude that the other point is a global maximum because the constraint gives us:

\(\displaystyle y=\frac{1}{2x}-2x\)

and substitution into the objective function gives us:

\(\displaystyle f(x)=28x^2+4x-7+\frac{4x+7}{4x^2}\)

which is unbounded as \(\displaystyle x\to\pm\infty\)

If we differentiate this function, and equate the result to zero, we obtain:

\(\displaystyle f'(x)=(2x+1)(2x-1)\left(28x^2+2x+7 \right)=0\)

As before, the yields the critical values:

\(\displaystyle x=\pm\frac{1}{2}\)

Now, the second derivative of the objective function in $x$ is:

\(\displaystyle f''(x)=448x^3+24x^2-2\)

And we find that:

\(\displaystyle f''\left(-\frac{1}{2} \right)<0\)

\(\displaystyle f''\left(\frac{1}{2} \right)>0\)

And so we know there is a relative minimum and a relative maximum, but given the behavior of the function at the extremes, i.e.:

\(\displaystyle \lim_{x\to\pm\infty}f(x)=\infty\)

We must therefore conclude that the relative minimum is the global minimum while the relative maximum is not the global maximum.