Find the minimum and maximum values of x+2y where x and y are lengths in the figure attached inside and 0≤x≤10. **write EXACT answers*** minimum=__________. maximum=___________.
figure! :D
Is the figure a "closed" figure?
umm not sure.. it doesn't specify :/ do you mean like an interval? :/
Is this calculus?
yes:)
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\[ y^2 = 5^2 + (10-x)^2 \]
okay:) using pythagorean theorem right?
\[ f(x) = x + 2y = x + 2\sqrt{5^2+(10-x^2)} \]
oh so from here, do we need to find the derivative? :/
We need to find critical points. Any place where \(f'(x)\) is \(0\) or undefined.
In this case \(x=0\) and \(x=10\) are critical points as well, simply because they are on the boundary.
ohh okay:) so here we plug into the original function? f(0)= f(10)= ?
or is it a different step? haha not sure if i mixed up the concepts :/
We are only using \(f'(x)\) to find critical points. Once we have all the critical points, we plug them all into \(f(x)\) to find the max and minimum values.
okay so f(0)=x+2y = 0+2y = 2y? is it like that?
No, I already told you...
\[ f(x) = x + 2y(x) = x+2\sqrt{5^2+(10-x)^2} \]
In short: \[ f(x) = x+2\sqrt{5^2+(10-x)^2} \]
First find \(f'(x)\). Then find \(f'(x) = 0\) and \(f'(x)\) is undefined.
ohh okay oops.. my bad!! so \[f(0)=0+2\sqrt{25+(10-0)^2 }\] which equals \[f(0)=0+2\sqrt{125}\] ?
oh derivative first, then plug in?
Yes, derivative first. Find all critical points first. Once you have all the critical points, you can plug them into \(f(x)\).
umm so would the derivative be \[f ' (x) = 0 + \frac{ 2(x-10) }{ \sqrt{x^2-20x+125} }\] ?
is that what i would be plugging in 0 and 10 into?
\[ \frac{d}{dx}x = 1 \]
\[ \frac{d}{dx}2\sqrt{5^2+(10-x)^2} = 2\frac{d}{dx}\sqrt{x^2-20x+125} \]
Ummm, you can do the rest I think.
ahh not sure what's happening her :/ is it implicit differentiation?
*here
@wio ? :/
sorry, deriving that part is really confusing me :(
No, it isn't implicit because I already gave you an explicit function.
This is the function\[ f(x) = x+2\sqrt{5^2+(10-x)^2} \]You just have to differentiate this function with respect to \(x\).
ohh okay, thank you!!:)
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