Question

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Pre-Calculus; story-math quadratic functions
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Answer 1

I believe the square sides formed from the x would be\[\frac{x}{4}\]in length, and the length of the sides for the square formed from the 10-x would be\[\frac{10-x}{4}\]

Answer 2

Well, you know the perimeters will be \(x\) and \(10-x\). So you need to first find the side length, and then square it.

Answer 3

Yes, I did that. I'm not sure if it's right though. @wio

Answer 4

So based on the previous comment of mine:\[A_{small}=(\frac{x}{4})^2=\frac{x^{2}}{16}\]\[A_{large}=(\frac{10-x}{4})^{2}=\frac{(10-x)^{2}}{16}=\frac{2x^{2}-20x+100}{16}\]

Answer 5

How am I doing so far you guys? ○ v ○

Answer 6

the area of the 1st one is ok
on the 2nd one, you wouldn't get a \(2x^2\) but just an \(x^2\)

Answer 7

Oh yeah, sorry. I got that after adding the two areas (I copied my work down wrong lol)

Answer 8

well, then you're doing good :)

as of course, to find how much area they both enclose, simply ADD them up
and to mininize their enclosure, simply SUBTRACT them

Answer 9

Eh? • - •
How does subtracting minimize it? lol

Answer 10

hmmm maybe I misread one sec =)

Answer 11

I think it's finding the (local?) minimum or vertex?\[h=-\frac{b}{2a}\]

Answer 12

I think you're right, yes

Answer 13

either way.. you're doing good

Answer 14

both quadratics, have a positive leading term coefficient, thus, the parabola opens up
thus the vertex is the lowest point for "y" and that'll be the lowest you can get for the area

Answer 15

Yeah. Global minimum, I mean ...

So then I tried to simplify the numerator:\[\frac{2x^{2}-20x+100}{16}\rightarrow\text{ numerator: }2x^{2}-20x+100=2(x^{2}-10x+50)\]Does this work?

Answer 16

should, yes

Answer 17

well.. wait a sec

Answer 18

hmm you're still using \(\bf 2x^2\) though, whcih will help with the simplifying, but sadly is only \(\bf x^2\)

Answer 19

I get \[\frac{2(x^{2}-10x+50)}{16}=\frac{x^{2}-10x+50}{8}\]Wait what?? D:

Answer 20

you don't have a \(\bf 2x^2\) to use for the common factor, is \(\bf x^2\)

Answer 21

???\[\frac{x^{2}}{16}+\frac{x^{2}-20x+100}{16}=\frac{x^{2}+x^{2}-20x+100}{16}\]

Answer 22

That gives me 2x^2 as the leading coefficient right x_x

Answer 23

ohh. though you were doing the 2nd part, ok

Answer 24

hmmm rats.. misread again... I see what you're doing, doing the global minimum, thus the \(2x^2\)

Answer 25

OH! Hahah I'm so sorry, I didn't make it clear enough > <;;

Answer 26

ok... that looks fine

Answer 27

Ok... but I tried to quadratic function the numerator and got an unreal sqrt o_e

Answer 28

\(\bf \cfrac{2x^2-20x+100}{16}\implies \cfrac{1}{8}x^2-\cfrac{5}{4}x+\cfrac{25}{4}\)

Answer 29

\[x^{2}-10x+50\rightarrow x=\frac{-(-10)\pm\sqrt{(-10)^{2}-4(1)(50)}}{2(1)}=\frac{10\pm\sqrt{100-200}}{2}\]

Answer 30

I -- what?! o-e how?

Answer 31

heheeh, don't forget the denominator

Answer 32

Well yes but I am trying to factor this numerator first lol\[x^{2}-10x+50\]

Answer 33

Should I just remove the factoring out of the 2 and factor this one instead?\[2x^{2}-20x+100\]

Answer 34

(because I messed up after factoring out a 2)

I will try using the Quadratic Formula on the original polynomial...

Answer 35

well... pardon my ignorance, why do you need the factoring? :) thought you were looking for the x-coordinate of the vertex

Answer 36

but how do I simplify this D:\[\frac{2x^{2}-20x+100}{16}\]

Answer 37

\(\bf \cfrac{2x^2-20x+100}{16}\implies \cfrac{1}{8}x^2-\cfrac{5}{4}x+\cfrac{25}{4}\) :)

Answer 38

I don't know how you got that > <;;

Answer 39

by distributing the denomiantor :)

Answer 40

denominator rather, darn typo =)

Answer 41

AH
So like this?\[\frac{2x^{2}}{16}-\frac{20}{16}+\frac{100}{16}\]

Answer 42

lol denomiantor haha xD

Answer 43

\(\cfrac{2x^2-20x+100}{16}\implies
\cfrac{2}{16}x^2-\cfrac{20}{100}x+\cfrac{100}{16}
\implies
\cfrac{1}{8}x^2-\cfrac{5}{4}x+\cfrac{25}{4}\)

yeap

Answer 44

Ohh, ok. I see :-)

Answer 45

from there, you can get "x" by -b/(4a)

Answer 46

so I get...
2/16 becomes 1/8, -20/16 becomes 5/4, and 100/16 is 25/4

Answer 47

then I use -b/2a...
WAIT. 4a ??????????

Answer 48

Sorry I hit the ? key too hard lol

Answer 49

ohh shoot, 2a yes.. got the y sticking there anyhow.. =)

Answer 50

Ok, yes thank you > <
Hmm\[-\frac{b}{2a}=-\frac{-\frac{5}{4}}{2\times\frac{1}{8}}\]

Answer 51

Good so far?

Answer 52

\(
{\color{red}{ \cfrac{1}{8} }}x^2{\color{blue}{ -\cfrac{5}{4 }}}x{\color{green}{ +\cfrac{25}{4} }}
\\ \quad \\
\textit{vertex of a parabola} \quad


\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\)

Answer 53

yes

Answer 54

I thought it was this lol\[-\frac{b}{2a},f(-\frac{b}{2a})\]

Answer 55

is both

Answer 56

you can use either to get "y"

Answer 57

Ah okay, the teacher said use the one I mentioned so... x_x

Answer 58

\(\textit{vertex of a parabola}\\ \quad \\
y = {\color{red}{ a}}x^2{\color{blue}{ +b}}x{\color{green}{ +c}}\qquad

\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)\\ \quad \\

% vertex of a vertical parabola, using f(x) for "y"
\textit{or as }\qquad \qquad \qquad \quad
\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad f\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\right)\right)\\ \quad \\

% vertex of a horizonal parabola, using f(y) for "x"
x = {\color{red}{ a}}y^2{\color{blue}{ +b}}y+{\color{green}{ +c}}\qquad

\left(f\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\right)\quad ,\quad -\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}} \right)\)

in case you want to write them down =)

Answer 59

Ok so I get...\[\frac{-\frac{5}{4}}{\frac{2}{8}}=-\frac{5}{4}\times\frac{8}{2}\]Good so far? @jdoe0001

Answer 60

yes

Answer 61

Alright (the "Viewing" bar only showed my icon, sorry D: )...
then I get this -\[-\frac{5\times 8}{4\times 2}\rightarrow-\frac{40}{8}\rightarrow -5?\]

Answer 62

yes, kinda
recall \(\bf -\cfrac{b}{2a}\iff -1\left( \cfrac{b}{2a} \right)\implies -1(-5)\implies +5\)

Answer 63

Dern OS keeps kicking my icon off the "Viewing" bar = =;
Oh?

Answer 64

\(\bf {\color{red}{ \cfrac{1}{8} }}x^2{\color{blue}{ -\cfrac{5}{4 }}}x{\color{green}{ +\cfrac{25}{4} }}
\\ \quad \\
\textit{vertex of a parabola}\qquad


\left(-\cfrac{{\color{blue}{ b}}}{2{\color{red}{ a}}}\quad ,\quad {\color{green}{ c}}-\cfrac{{\color{blue}{ b}}^2}{4{\color{red}{ a}}}\right)
\\ \quad \\ \quad \\
thus
\\ \quad \\
-\left( \cfrac{-\frac{5}{4}}{\cancel{2}\cdot \frac{1}{\cancel{8}}} \right)\implies \cfrac{\frac{5}{4}}{\frac{1}{4}}\implies \cfrac{5}{\cancel{4}}\cdot \cfrac{\cancel{4}}{1}\implies 5\)

Answer 65

Oh yeah, I forgot about the double negative...
Oops. ^-^;

Answer 66

so x =5, be happy, eat ice-cream, do hula-hoop

Answer 67

xD

Answer 68

I will!! Thank youu \ ( ^ ○ ^ \)

Answer 69

yw

Answer 70

can I have a medal? > w >
lol > <