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I have worked on this one for a while and I can not make my answer match the author's.

Find the points on the ellipse $4x^2 + y^2 = 4$ that are the farthest away from the point (1,0).

I have:

$$4x^2 + y^2 = 4$$

and then the distance formula, so I set y to terms of x and I get

$$\sqrt{(x-1)^2 + (2-2x)^2}$$

Setting the difference the a square this gives me a derivative of

$$10x-10$$

which gives me a zero of 1, this is wrong according to the book and I am not sure why.

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    Of course it gives you that the minimum is when $x = 1$. You've just said that the point closest to $(1,0)$ is, in fact, $(1,0)$! But you haven't used that the points need to be on an ellipse.2012-04-04
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    Did you do $\sqrt{4-4x^2}=\sqrt4-\sqrt{4x^2}$? Please tell me you didn't do that....2012-04-04

1 Answers 1

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We want to maximize $\sqrt{(x-1)^2+(y-0)^2}$, given that $4x^2+y^2=4$.

Equivalently, we want to maximize $(x-1)^2+y^2$, same side condition.

But $y^2=4-4x^2$.

So we want to maximize $(x-1)^2+(4-4x^2)$. Your turn.

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    To expand on Andre's "equivalently": Since the square root function is increasing everywhere, the $x$ and $y$ that maximize the square root of "something" will also maximize the "something". Once you have this $x$ and $y$, however, it's important to remember to plug them back into the *original* equation (the one with the square root) to determine the maximum itself.2012-04-04
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    Why am I not maximizing a square root, or am I squaring the max?2012-04-04
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    If you like, work with the square root. The derivative will just look more complicated.2012-04-04
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    I guess I don't understand what I did wrong, this gave me the correct answer but I am not sure how it is different from what I did. Actually taking the derivative of my original equation is wrong and I am not sure why. I get 10x-10.2012-04-04
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    Jordan, I'm not sure where you got $$y = 2-2x^{2}.$$ You should have gotten $$y^{2} = 4-4x^{2}$$ as shown by the original equation. Perhaps you should should note $(2-2x^{2})^{2}$ is not equal to $4-4x^{2}$2012-04-04
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    I thought $y^2=2-2x^2$ so I just took the square root of that which I do not see as being wrong because I can't take the square of a negative.2012-04-04