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The height of triangle ACE is 12 and base AE = 20. A rectangle is inscribed in triangle ACE as shown below.? a) Express the rectangle’s area as a function of x and y. Use the fact that the triangle ACE is similar to triangle BCD to write y as a function of x. c)Express the rectangle’s area as a function of only x. d) Find the maximum possible area of the rectangle.

Mathematics
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|dw:1345995812545:dw|
Nobody knows what to do?:#

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Other answers:

a) (12-y)x b) 12(x-0.1x^2) c) 30 sq units
...How?
Is "How" for all of it?
Just b and c
especially b... :O
consider the perpendicular line form C, we can come out with the ratio 12/y =10/x, where 10 is half the rectangles side,20. Then simplify it so as to inset the result in to the equation in 'a'. To 'c' just use the application of derivative.
Okay. Makes sense I guess:p thanks!
small catch
triangle ACE is similar to triangle BCD => height1/height2 = base1/base2
12/y = 20/x y = (3/5)x
this would give you max area 60
o.O But y needs to be written as a function of x.
y = (3/5)x
its a function of x
linear
Oh yeah :3. So why did you use the entire base and @Zekarias used 1/2?
since both are similar triangles, "ratio of bases" equals "ratio of heights"
Okay, that makes sense. :3 Can you explain how you got 60 though?
write the area equation in x and optimize
for optimizing, use either of the below two : 1) its a parabola facing down so max value = f(-b/2a) or 2) use calculus
this q from conics, or calculus ?
calculus. Going into calculus next year and they gave us some prep stuff*.*
oh ok, so we have this so far : A = x(12-y) -------(1) y = (3/5)x --------(2)
Okay, true.
hmm which one moves ?
ohhk yeah as we go down its moving a lot it seems...
substitute (2) in (1)
x(12-(3/5)x)
\(A = x(12 - \frac{3}{5}x)\) \(A = 12x- \frac{3}{5}x^2)\)
Okayy and solve for x
nope we need to optimize A
find maximum possible value of A, right ?
Oh.
Ya
theres a process for finding maximum/minimum values of any funcitons
you familiar with differentiation ?
... I guess I havenet learned that yet..
its simple... if you knw differentiation already
I think we are learning that next year :( (monday)
just if you knw derivative of x, and x^2, its enough
oh ok then you cannot use calculus, lets switch to conics. you knw parabolas, right
Okay, I don't know how to do that.. Maybe I should make a new question asking for the dirivative of A=12x-3/4x^2
Yes.
You said f(-b/2a)
yes
A = 12x - (3/5) x^2
\(A = \frac{-3}{5}x^2 + 12x\)
Yes
a = -3/5 b = 12
Cool, thanks, just plug and chug now:D
f(-b/2a) gives the vertex y co-ordinate, which is the maximum value
y=10
|dw:1345999525420:dw|
-b/2a = 10 f(-b/2a) = ?
put x = -b/2a = 10, in area function
Okay.
60!:D
if you remember, for any parabola, -b/2a is the x-coordinate of vertex f(-b/2a) is the y-coordinate of vertex
great !!
Thank you so much! Now I can do the rest of these problems:D
glad to hear ! calculus wud be much fun :) good luck !!

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