24x36" picture frame has a uniform width around it. The area is 396. What is the width?

Question

a 24 in by 36 in. picture has a frame of uniform width around it. the area of the frame is 396 in^2. What is the width of the frame?

Answer 1

Strategy:

The key is to draw the picture and its frame, and label everything.

The most common mistake is to double-count or under-count the corners.

The greatest challenge to the problem is remembering how to find a solution to a quadratic equation.

You should memorize the quadratic formula.

Even better, be able to derive the quadratic formula. Although you will not have time (during a test or homework) to derive the formula, nevertheless, knowing its derivation will allow you to avoid errors of memory. Deriving the quadratic formula is a simple but lengthy exercise that is well worth the time.

Step-by-step solution:

First, draw a diagram / picture

The original picture (before framing) has dimensions 24 x 36 in²

XXX

XXX

Next, draw the width of the frame.

In every "frame" problem, the most common error is to double-count or under-count the corners.

You are adding a bit to each side

xxxxxx

xXXXx

xXXXx

xxxxxx

Don't double-count or under-count the corners. How? ... Be careful, like this:

Count each width separately:

The top, in our picture, will have an area, A_l = (width) * 36 in²

The bottom will have the same area A_l = (width) * 36 in²

Each side will have an area, A_s = (width) * 24 in²

A_s = (width) * 24 in²

Are we finished? ... What about the corners?

... Aha! We have to add one more thing: four corners, each having an area, A_c = (width) * (width) in²

Since we know that the total area of the frame (not picture) = 396 in²

Then we can set up an equation like this:

( A_s + A_s + A_l + A_l + 4A_c ) in² = 396 in²

Which we can expand by substituting our values from above:

(2 A_s + 2 A_l + 4 A_c) in² = 396 in²

(2*24*w) + (2*36*w) + (4*w*w) = 396 in²

And now you have to solve it as a quadratic equation (because there is a squared variable w):

4w² + 120w - 396 = 0 (units in in²)

Simplify (dividing each term by coefficient 4)

w² + 30w - 99 = 0

Now can you solve for w?

There are three ways to solve any quadratic:

1) Factor directly

2) Complete the square

3) Use the quadratic formula (which is just a rearrangement of the standard formula, solved for x, which is w in our equation)

Memorize this formula!

w = { (-b) +/- [Sqrt (b² - 4ac)] } / 2a

where a, b, and c are the coefficients for the quadratic equation in standard form, that looks like this:

aw² + bw + c = 0

See from our equation

w² + 30w - 99 = 0

that a = 1, b = 30, and c = (-99)

Then, plug in those values to solve for w, using the quadratic formula:

w = { (-b) +/- [Sqrt (b² - 4ac)] } / 2a

This gives two real solutions, w = -38 and w = 3

(Side note: There will always be 2 solutions, but they may be real or complex. For example, if you need to take the square root of a negative number, then you will have a complex solution, not a real solution. When you are solving a problem with a familiar geometry like this one, you can safely ignore complex solutions.)

Since we can't have a negative in a geometric solution (that would be called the degenerate form), therefore, the w= -38 doesn't work. We have only one solution that works for this geometry: 3

Now, as always, we check the solution to verify that it works. Here, our solution is w = 3. Does this check out?

?? 396 = (2*24*3) + (2*36*3) + (4*3*3)

?? 396 = 396 (Check)

Yes, this checks out. The equation works for us, when w = 3. That means our solution is correct: w = 3.

Therefore, the width of the frame is 3 inches (Solution)