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You know the steps and notation so I am going omit much of the notation for simplicity:

1) $8=8x\lambda$

2) $8=8y\lambda$

3) $5=10z\lambda$

multiplying equations 1 and 2 by 5 and equation 3 by 8 gives and diving out by 8 gives:

$x=y=2z$

Now replace $x$ and $y$ with $2z$ plug them into the constraint:

$z=\pm1$

now since put $\frac{1}{2}x$ or $y$ (since the are the same) in for z and solve for x:

$4x^2+4y^2+\frac{5}{2}x^2=37$

$x=y= \pm \frac{\sqrt{74}}{\sqrt{21}}$

I just don't think something is right about my x and y's given that these are functions from a textbook, answers like these are usually a red flag, I have a final in two days. I know thata after this step I take all my triplets and plug them into f(x) to find the maximum and minimum values from the function

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$$x=y=2z$$

If $z=\pm 1$

Then we should have $x=y=2z=\pm 2$

A mistake is

$$4x^2+4y^2+\frac52 x^2 = 37$$

where it should have been $$4x^2+4y^2+\frac54 x^2 = 37$$

where you have forgotten to square the $\frac12$.

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  • $\begingroup$ Okay very clear where my mistake was thankyou $\endgroup$ – K. Gibson Dec 2 '16 at 19:19

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