SOLUTION: The area of a 6 m by 4 m rectangle is to be doubled. By what equal amount must each dimension be increased?

Hi, there--

The problem:
The area of a 6 m by 4 m rectangle is to be doubled. By what equal amount must each dimension be increased?

A solution:
Let x be the amount by which each dimension is increased.
Then the side lengths of the larger rectangle are 6+x and 4+x meters.

The area of the smaller rectangle is 6*4=24 square meters.
The area of the larger rectangle is doubled, so its area is 24*2=48 square meters.

Now we use the formula for the area of a rectangle (length times width equals area) to write an equation and solve for x.
(6+x)*(4+x) = 48

Use the distributive property to simplify the left side.
24+6x+4x+x^2 = 48

Rearrange and combine terms.
x^2 + 10x + 24 = 48
x^2 + 10x - 24 = 0

Factor. You want two factors whose sum is 10 and whose product is -24. The factors are -2 and 12.
(x + 12)(x - 2) = 0
x + 12 = 0 OR x - 2 = 0
x = -12 or x = 2

x = -12 has no meaning in this problem since the rectangle cannot have negative dimensions.
x = 2 means that we increase each side length by 2 meters.

Check.
The side lengths of the larger rectangle are 6+2=8 and 4+2=6. The area of the larger rectangle is 8*6=48. The area of the large rectangle is double the area of the smaller one 
since 48 is double 24.

That's it. Feel free to email me if you have questions about the solution.

Ms.Figge
math.in.the.vortex@gmail.com