Divide 40 into two parts such that the sum of the squares of these parts is a minimum.

Question

gorv · Answer

hmmm..let one part is x

blackbird02 · Answer

ok, what to do next?

anonymous · Answer

$$x^2+y^2$$

x = equals one side however the otherside = 40 - x


$$x^2+(x-4)^2 = y$$

anonymous · Answer

wait lol

$$x^2+(40-x)^2= y$$

gorv · Answer

now use differentiation to find least  or max value

anonymous · Answer

you there?

anonymous · Answer

I have the answer but i need to see you're working towards getting the answer yourself

blackbird02 · Answer

so it's gonna be 2x^2+1600-80x=y

anonymous · Answer

not quite

anonymous · Answer

how'd you get 1600 - 80 x

anonymous · Answer

derivative rule

$$x^n = nx $$

blackbird02 · Answer

i simplified (40-x)^2

anonymous · Answer

and the other is

$$(x+c)^n=n(x+c)(x+c)'$$

anonymous · Answer

ehh you can do it that way but if you just use the rule that i showed you... it comes out to be really simple

anonymous · Answer

$$2x +2(40-x)(-1) = y'$$
$$2x+(80-2x)(-1) = 2x - 80 +2x = 4x-80$$

$$4x - 80 = y '$$

anonymous · Answer

but it looks right if you factor out everything

anonymous · Answer

so you just take the derivative of that and you'll get what i got, correct?

anonymous · Answer

simple rules ... you only need the first rule when factoring out.... anyways, lastly, what does your book say about y' = 0

anonymous · Answer

@blackbird02

blackbird02 · Answer

there wasn't anything stated about it

anonymous · Answer

alright well think of it this way... y' or your derivative is the slope of your equation  correct?

anonymous · Answer

well y' is a tangent line with the same slope as your line... therefore if you have say...... x^2 as your equation...

anonymous · Answer

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