# algebra maximum area picture frame

• May 11th 2005, 04:52 AM
ahsas23
algebra
A picture framer has a piece of wood that measures 1 inch wide x 50 inches long with which she would like to make a picture frame with largest possible interior area.

1. Write an equation in x andy for outer perimeter of the frame.

2. Find the largest interior area of the picture frame possible.

3. Find the external dimensions of the picture frame that provide the largest interior area of the frame.

4. Do 2 and 3 algebraically and graphically

The instructions doesnt state a measurement for the depth,
• May 11th 2005, 07:40 AM
paultwang
What is the depth?
• May 11th 2005, 07:43 AM
Math Help
This is a two dimensional problem. You should not need depth.
• May 12th 2005, 03:52 AM
ticbol
If you want this solved by Algebra only, not by Calculus, then we need to use Analytic Geometry. Is Analytic Geometry a part of Algebra?

Anyway, by Analytic Geometry I mean by using the properties of a parabola, a vertical parabola that opens downward. The vertex of this parabola is the highest point of the curve, and that shows the value of the maximum area.

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1. Write an equation in x and y for outer perimeter of the frame.

Let x = exterior or outer width of the frame, in inches.
And y = exterior height, in inches.

Exterior perimeter, P = x +y +x +y = 2x +2y = 2(x+y). ---answer.

If for maximum P,
max P = 2(x+y) = 50
x +y = 50/2
x +y = 25 ------(1)

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2. Find the largest interior area of the picture frame possible.

Since the width of the piece of wood is 1 inch, then,
Interior width = x -1 -1 = (x -2) in.
Interior height = y -1 -1 = (y -2) in.
So,
Interior Area, A = (x-2)(y-2) sq.in.

For maximum A, we use the Eq.(1) above.
Solving for y in Eq.(1),
x +y = 25 ------(1)
y = 25 -x

A = (x-2)(y-2)
A = (x -2)[(25 -x) -2]
A = (x -2)(23 -x)
A = 23x -x^2 -46 +2x
A = -x^2 +25x -46 ----a vertical parabola that opens downward.

Using formulas, the x-coordinate of the vertex is
x = -b/(2a)
So, the x-coordinate is -25/(2 * -1) = -25 / -2 = 12.5

Then, the A-coordinate of the vertex, which is the maximum A, is
A = -x^2 +25x -46
max A = -(12.5)^2 +25(12.5) -46
max A = -156.25 +312.5 -46
max A = 110.25 sq.in. ----answer.

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3. Find the external dimensions of the picture frame that provide the largest interior area of the frame.

Exterior width = x
max exterior width = 12.5 in.

Exterior height = y = 25 -x
max exterior height = 25 -12.5 = 12.5 in. also.

Therefore, for largest interior area, the frame is 12.5 inches by 12.5 inches on the outside. ...answer.

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4. Do 2 and 3 algebraically and graphically

I cannot show it graphically here. I don't know how to draw it here.