Question

calculus help
http://prntscr.com/9lo0p8

Answer 1

So, here we'll have to work with two rectangles.
The big one (b), whose maximum perimeter is 40ft, and the one defined inside the big rectangle (b) we will call (s) and is bounded by the house.
So, we will then call the with and the length of rectangle (b) as "x" and "y" respectively, therefore, the area and perimeter of (b) will be defined as:
\[A _{(b)}=x.y\]
\[P _{(b)}=2(x+y)\]
Now, this would suffice if the rectangle in consideration was not bounded by (a). so, all we have to do, is sustract from (b), the perimeter that coincides with (b) and the area of (a) from (b), this will give us the figure (c), then:
\[A_c=A _{(b)}-A _{(a)}\]
\[P_c=P_{(b)}-(6+4)\]
And those are the values that will allow us to calculate and form a function which will allow us to optimize the value of the given fencing:
\[A_c=(x.y)-(6.4) \iff A_c = x.y-24\]
\[P_c=2(x+y)-10\]
Now, we know that the maximum value of fencing will be 40, so therefore Pc=40:
\[P_c=40 \iff 2(x+y)-10=40\]
Look at the area Ac is dependant of two variables x and y, we don't want that, we want a function that is only dependant of "x", so we will then, from the last equality solve for "y":
\[2(x+y)-10=40 \iff 2x+2y=50 \iff y=\frac{ 50-2x }{ 2 } \iff y=25-x\]
Now, we substitute the value of "y" we found on Ac:
\[A_c=(x.y)-24 \iff A_c(x)=(x(25-x))-24\]
Then the resulting function is:
\[A_c(x)=25x-x^2-24 \iff A_c(x)=-x^2+25x-24\]
Now, all you have to do is find the maximum or minimum points of the function:
\[A_c(x)=-x^2+25x-24\]

Answer 2

i take the derivative of that function to find the max or min right?

Answer 3

Yes.

Answer 4

x=12.5

Answer 5

Is it a maximum or a minimum?

Answer 6

a max i think

Answer 7

Why?

Answer 8

if the framing is only on the dashed line then your perimeter function is not correct

Answer 9

I considered the other end, since there is no specification for the last fragment.
And I don't think it is something we can find is it?

Answer 10

Hmm ya, why did you subtract 10 from the perimeter? :o
Thinking...

Answer 11

I agree with @Zarkon - only dashed line represents the framing

Answer 12

Oh oh oh, no frame touching the house?
I see, interesting.

Answer 13

I also thought the dashed line only did represent the fencing, but that be the case the final section would've been a given.

Answer 14

that length depends on the shape which varies with x

Answer 15

i.e., fixing that section would determine the area at once

Answer 16

I see the thing now.
The area would still be calculated by the difference of the smaller rectangle with the other, though the perimeter utilized will be:
\[P _{(b)}=(x+2y)\]
\[P_c=(x+2y)-4\]

Answer 17

\[P_c=40 \iff (x+2y)-4=40 \iff y=\frac{ 44-x }{ 2 }\]
\[A_c(x)=(x(22-\frac{ x }{ 2 }))-24\]
\[A_c(x)=-\frac{ x^2 }{ 2 }+22x-24\]
Correct me if I'm wrong.

Answer 18

Ahh darn :) I did mine in y, not x lol
harder to compare

Answer 19

I got x=22, a max

Answer 20

is that the answer?

Answer 21

It seems to be so, I find no problems aside from my initial mistake.

Answer 22

That's not the "answer",
but yes you're on the right track! :)

That's the x value which maximizes your area,
they still want the area amount.

Answer 23

Don't forget about the units.

Answer 24

do i plug 22 into -x^2/2+22x-24?

Answer 25

Yes.

Answer 26

i got 702

Answer 27

mm, does not seem correct.

Answer 28

Did you make sure to plug it in correctly?:
\[A_c(22)=-\frac{ 22^2 }{ 2 }+22(22)-24\]

Answer 29

218

Answer 30

That's correct, now to finally write the answer, remember to write it in its corresponding unit.

Answer 31

218 ft sq

Answer 32

\(A_c\)=218 ___
Should be enough.

Answer 33

Thank you so much for the help!

Answer 34

No problem, and I'm sorry for my initial mistake there.

Answer 35

That's ok :)