a rectangular picture 8 inches by 12 inches has a frame of uniform width x inches

Question

A) write a function expressing the total A of the picture and the frame as a function of x.

B) write your equation in vertex form. What are the coordinates of the vertex?

C) find the total area when the width of the frame is 3 inches.

D) Suppose you want the total area of the frame and the picture to be twice the are of the picture. Solve your equation for x

E) How wide should the frame be if the total area is twice the area of the picture?

F) now suppose you want the total are to be fives times the area of the picture. Is the width of the frame five times your answer to part (e)? Explain.

Answer 1

What a doozy! Make sure you draw a picture for this one!

 

The area of the picture itself is 8×12. Draw this rectangle.

 

Now there is a frame around the outside of it which is x inches wide. Label it.

So how wide is the entire frame AND picture? Well for the width, it's 8 inches across the picture, then there's an x on the right side and an x on the left side. Looks like it's 8+2x inches wide. Same deal with the height, it looks like it's 12+2x inches high. Since Area=Height×Width, we should have our function.

 

A) A=8×12=96      A(x)=(8+2x)(12+2x)

 

We can distribute out these binomials a bit so that it actually feels like a quadratic. my expanded form looks like:

 

A(x)=4x²+40x+96.

 

Converting to vertex form is a lot like completing the square: I want to find a binomial I can square.

 

First I want to factor our a GCF of 4:

 

A(x)=4(x²+10x+24)

 

My method is to add and then subtract half of my middle coefficient. In this case, half of 10 squared is 25

 

A(x)=4(x²+10x+25-25+24)

 

Now, I remember that when I square a binomial I get (a+b)²=a²+2ab+b².

 

 

 

A(x)=4((x²+10x+25)-25+24)

 

What I have grouped in the inner parentheses here is a squared binomial, so we can write it as (a+b)² and simplify everything else

 

A(x)=4((x+5)²-1)

 

Did you follow that? If not, ask another question about completing the square.

 

I want this in the vertex form: a(x-h)²+k, so I should probably distribute that 4 to the -1, and then I'll have it in vertex form.

 

B) A(x)=4(x+5)²-4

    vertex: (-5, -4)

 

Finding the area when x=3 is as simple as plugging in 3 for x in our function.

A(3)=4(3+5)²-4

A(3)=4(64)-4

 

C)  A(3)=252 in²

 

If we want the total area to be twice the area of the picture, we should start by figuring out what area that is. Since the are of the picture is 96. Twice that is 192. So when is the area 192?

 

A(x)=192

192=4(x+5)²-4

 

Let's get that squared binomial alone:

 

196=4(x+5)²

49=(x+5)²

x+5=±7

x=-5±7

so x is -12 OR 2.

 

 

Since x is a distance, it had better be positive, so we can ignore one of our answers.

D) x must be equal to 2

 

Since our function A(x)=4(x+5)²-4 is the area of the frame AND picture, the frame alone would be our function minus the picture, or 4(x+5)²-4-96=4(x+5)²-100

So when is this equal to twice the picture?

192=4(x+5)²-100

Again isolate our squared binomial.

292=4(x+5)²

73=(x+5)²

±√73=x+5

x=-5±√73

Again, x is a distance, so we ignore our negative answer.

E) x=-5+√73≈3.5