Question

Can anyone please help me with this expression: (x-2)((L/2)-x-2)? I am trying to solve for L. I cannot seem to do it though, is there no solution for just L? (I am trying to isolate x in another expression by plugging in the solution of L, so that I can find x)

Answer 1

Also, I am really not sure about what this actually means: "Use your answer to part (B) to write an expression (i.e. a function) of its exterior height x, and the wood length L." Does it mean that I am supposed to write a function of BOTH x and L, or a function of x, with L in it?

Answer 2

(x-2)((L/2)-x-2) is a expression, you cannot solve it (there's nothing to solve). can you post the complete question?

Answer 3

It is sort of a series of questions, but basically, I am trying to solve this: "(A) Write an equation that expressed the required relationship between variables x,y, and the total length of the wood L. In Part 1, you had a linear equation of the form ax_by=50 where a and b were numbers. Now you will have a slightly more complicated equation involving x, y, and L, i.e. your equation will have three variables, x, y, and L in it.

Answer 4

So I got 2x+2y=L for that.

Answer 5

And then the second part is "(B) Solve your equation from part (A) in terms of x and L."

Answer 6

I got y=(L/2)-x for that.

Answer 7

And then this is the final part that I am stuck on: "(C) Use your answer to part (B) to write an expression (i.e. a function) of its exterior height x, and the wood length L. [The variable y will not appear in this equation]"

Answer 8

And I am pretty sure I am supposed to solve for L, in that expression.

Answer 9

I am basically trying to get L=[something] to plug into the original expression/function (x-2)((L/2)-x-2), to have only one variable, x, if that makes any sense.

Answer 10

it seems that you got (x-2)(L/2 -x -2) from (x-2)(y-2) what does this expression say physically?

Answer 11

Well you see, I got (x-2)(L/2-x-2) from (x-2)(y-2) because I plugged in my solution for y=(L/2)-x into the y, as there is not supposed to be a variable y (x is the height, y is the width).

Answer 12

what I am asking is what does this expression mean. That is is this an area or something

Answer 13

Oh, yes, it is the interior area of a picture frame.

Answer 14

There was a first part, it is relatively simple as it gives you the length, but it is basically just finding the interior area of a hypothetical picture frame, where each part of the frame is an inch wide, and you form the frame in various ways depending on the Length.

Answer 15

The reason why I am trying to solve this expression is so that I can plug in values of L that it gives me later on.

Answer 16

Maybe I am not actually supposed to do it this way, I am not sure...

Answer 17

ok. So, Is the question asking the values of x, y so that you'd get the maximum area?

Answer 18

Yeah, if it helps, the next question is "(D) Use your function to find the inside dimensions that will make this area a maximum. Now each of your inside dimensions will depend on L, i.e. they will be functions of L."

Answer 19

what you have to do is finding the length and the height of the frame in terms of "L" so that you'll get the maximum area for the frame.

take the area \[A=(x-2)\left(\frac{L}{2} -x -2\right)= -x^2+\frac{L}{2}x -2\left(\frac{L}{2}-2\right)=-\left(x-\frac{L}{4}\right)^2+\frac{L^2}{16}-2\left(\frac{L}{2}-2\right)\]

so, for A to become maximum, \[\left(x-\frac{L}{4}\right)^2\] has to become minimum which is 0.

\[\begin{align*}\left(x-\frac{L}{4}\right)^2&=0\\ x&=\frac{L}{4}\end{align*}\]

and \[y=\frac{L}{2}-x=\frac{L}{4}\]

Answer 20

If you know calculus, you can use

\[\begin{align*}
A(x)&=-x^2+\frac{L}{2}x-2\left(\frac{L}{2}-2\right)\\
\frac{dA}{dx}&=-2x+\frac{L}{2}
\end{align*}\]
when A is maximum \[\begin{align*}\frac{dA}{dx}&=0\\ -2x+\frac{L}{2}&=0 \\ x&=\frac{L}{4}\end{align*}\]

you can find y as in the earlier post

Answer 21

I am sorry, I get what you are saying, but in the beginning part, why is it \[-x^2+L/2-2(L/2-2)\]?

Answer 22

Specifically, how did you get the 2(L/2-2)?

Answer 23

\[\begin{align*}(x-2)\left(\frac{L}{2}-x-2\right)&=(x-2)\left(-x+\left[\frac{L}{2}-2\right]\right)\\
&=-x^2+\left(\frac{L}{2}-2+2\right)x-2\left(\frac{L}{2}-2\right)\end{align*}\]

Answer 24

OH! I see, I totally forgot keeping those in parentheses, thank you.

Answer 25

no problem. you're mostly welcome :)

Answer 26

So how would I write these as a "function of L"? L(x)=[whatever makes it a maximum]?

Answer 27

And to get that, I need to fully factor the expression, then set it to equal zero, then find each of the two possible inside dimensions?

Answer 28

function of L means write x in terms of L or x(L) not L(x)

Answer 29

So how would I write that out? x(L)=?

Answer 30

we have already found it x(L) = L/4

Answer 31

That is the only possible dimension? There is not another one?

Answer 32

it is the only answer. x=y=L/4

Answer 33

I guess I am confused as to this "Now each of your inside dimensions will depend on L, i.e. they will be functions of L." There is only one inside dimension that will be a maximum?

Answer 34

not the dimension. It is the area that is maximized under the constrain of 2x+2y=L.

Answer 35

What are the dimensions, then, of "(D) Use your function to find the inside dimensions that will make this area a maximum. Now each of your inside dimensions will depend on L, i.e. they will be functions of L."?

Answer 36

One more thing, I am supposed to create a table of values that I get from plugging in whatever number into L, and I know that when L=50 the area must be 529.

Answer 37

Just using reasoning. So I need some function dimensions of L that will agree with that solution.

Answer 38

1. A picture framer has a piece of wood that measures one inch wide by 50
inches long. She would like to make a picture frame as shown below, and uses
all 50 inches of wood she has. Notice that x and y in our sketch below are the
outside dimensions of the frame.
NOTE: In your paper, you will need to reproduce the figure and rewrite all of the
questions as they appear in this paper.
(A) Write an equation that expresses the required relationship between variables
x, y, and the total length of the wood.
(B) Solve your equation from part (A) for y in terms of x.
(C) Write an expression (i.e. a function) that gives the interior area of the frame
as a function of its exterior height x.
(D) Find the values of x that make the interior area of the picture frame equal to
zero. Explain why this is the case in the context of the picture frame problem.
Draw the frames that correspond to these two extreme conditions.
(E) Use your function to find the inside dimensions that will make this area a
maximum. State also the maximum internal area of the picture frame.
PART II
2. Now the picture framer would like to rework this problem for a piece of wood of
arbitrary length L. [In PART I, L was 50 inches. Now L becomes a new variable
so that we can generalize what we did in PART I for a piece of wood of any
arbitrary length, L]
(A) Write an equation that expresses the required relationship between variables
x, y, and the total length of the wood L. In PART I, you had a linear equation of
the form ax+by = 50 where a and b were numbers. Now you will have a slightly
more complicated equation involving x, y and L, i.e. your equation will have three
variables, x, y, and L in it.
(B) Solve your equation from part (A) for y in terms of x and L.
(C) Use your answer to part (B) to write an expression (i.e. a function) that gives
the interior area of the frame as a function of its exterior height x, and the wood
length L. [The variable y will not appear in this equation]
(D) Use your function to find the inside dimensions that will make this area a
maximum. Now each of your inside dimensions will depend on L, i.e. they will be
functions of L.
(E) Use your expressions for the inside dimensions to calculate the maximum
area of each picture frame that can be built from each length of wood L in the
table below. The first row is the problem you solved in PART I of this paper.
Show at least one sample calculation for this part of the problem.
Length L in Inches Area A in square inches
50
80
120

Answer 39

It may actually be easier if you could just see the entire problem set. :P

Answer 40

I REALLY appreciate this, by the way.

Answer 41

since you found out side dimensions i.e. x=y=L/4 the inside dimensions can be achieved from (x-2) and (y-2) and the area would be (x-2)(y-2)=((L/4)-2)^2. how did you get 529

Answer 42

I just looked at everything I did so far, and I am honestly not sure how I got 529.

Answer 43

Wouldn't 23+23=46 (as 50-4=46) and then 23x23=529 be the maximum internal area? I subtracted 4 because of the combined frame width of 4 inches, and we are only trying to find the maximum internal area.

Answer 44

you cannot substitute 4 from 50,

take the lenght x and width y of the outside of the frame,

2x+2y=50

inner frame width would be (x-2) and (y-2). Then the length of the inner frame would be,

2(x-2)+2(y-2) not (x-2)+(y-2) then,

2(x-2)+2(y-2)=(2x+2y)-8=50-8 (not 50-4)