another optimization problem
- Thread starter zebra
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skeeter
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- Dec 15, 2005
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you want to minimize the distance between \(\displaystyle \L (x, x^2+1)\) and \(\displaystyle \L (3,1)\).
remember the distance formula?
\(\displaystyle \L d = \sqrt{(x-3)^2 + [(x^2+1)-1]^2}\)
to make it a bit easier, if you minimize \(\displaystyle \L d^2\), you also minimize \(\displaystyle \L d\).
remember the distance formula?
\(\displaystyle \L d = \sqrt{(x-3)^2 + [(x^2+1)-1]^2}\)
to make it a bit easier, if you minimize \(\displaystyle \L d^2\), you also minimize \(\displaystyle \L d\).
ok...so you find the derivative of d^2 and set that to zero, correct? well i did that yet i got stuck>>
d^2=(x-3)^2 + x^4
2d (dd/dt)= 2(x-3) + 4x^3
i cancelled out the twos and got
(dd/dt)= [(x-3) + 2x^3]/d
i set this to zero and therefore got rid of the d but i couldnt factor it to find x?
d^2=(x-3)^2 + x^4
2d (dd/dt)= 2(x-3) + 4x^3
i cancelled out the twos and got
(dd/dt)= [(x-3) + 2x^3]/d
i set this to zero and therefore got rid of the d but i couldnt factor it to find x?
skeeter
Elite Member
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- Dec 15, 2005
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- 3,215
let the square of the distance = D ...
\(\displaystyle D = (x-3)^2 + x^4\)
\(\displaystyle D' = 2(x-3) + 4x^3\)
\(\displaystyle 4x^3 + 2x - 6 = 0\)
by inspection, an obvious solution to this equation is x = 1. the other two roots are imaginary.
\(\displaystyle D = (x-3)^2 + x^4\)
\(\displaystyle D' = 2(x-3) + 4x^3\)
\(\displaystyle 4x^3 + 2x - 6 = 0\)
by inspection, an obvious solution to this equation is x = 1. the other two roots are imaginary.