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. So essentially the picture looks like a rectangle with the measurements 1 inch around all sides and it says that the length of the rectangle is Y and the width is X. Alright with that knowledge... A. Write an equation that expresses the required relationship between variables x, y, and the total length of the wood. Got 2(x+y)=50 B. Solve your equation from part (A) for y in terms of x. Got y=25-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. x^2+25x-50?? Confused on everything from here on out. 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. The second part of it goes like this: 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. Solution I A) 50 inches is the perimeter of the frame: 2x+2y= 50 II) A) 2x+2y= L b) solve for y C) The interior area is length * width A = (x-2)(y-2) since the frame is 1 inch on each side. Now plug in the y value from part B D) set the two factors equal to zero. This basically gives the domain, between these two values. E) set y\' = 0 to find the critical points within the domain. ( on part II, remember that L is a constant, not a variable) .

1. 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. So essentially
the picture looks like a rectangle with the measurements 1 inch around all sides and it says that
the length of the rectangle is Y and the width is X. Alright with that knowledge... A. Write an
equation that expresses the required relationship between variables x, y, and the total length of
the wood. Got 2(x+y)=50 B. Solve your equation from part (A) for y in terms of x. Got y=25-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. x^2+25x-50?? Confused on everything from here on out. 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. The second part of it goes
like this: 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.
Solution
I A) 50 inches is the perimeter of the frame: 2x+2y= 50
II) A) 2x+2y= L
b) solve for y
C) The interior area is length * width
A = (x-2)(y-2) since the frame is 1 inch on each side.
Now plug in the y value from part B

2. D) set the two factors equal to zero. This basically gives the domain, between these two values.
E) set y' = 0 to find the critical points within the domain. ( on part II, remember that L is a
constant, not a variable)