You are making a rectangular table. The area of the table should be 10 ft^2. You want the length of the table to be 1 ft shorter than twice its width. What should the dimensions of the table be?
Okay, so, I'm not sure how to put this into an equation. The example in the book is very different to this problem. Please help!!
A = LW
10 = W(2W - 1)
10 = 2 * 5
great!
To solve this problem, let's first define our variables:
Let's say the width of the table is 'w' ft (in feet).
According to the problem, the length of the table is 1 ft shorter than twice its width, so the length would be (2w - 1) ft.
We know that the area of a rectangle is equal to its length multiplied by its width.
So, we can set up the equation:
Area = Length * Width
10 ft^2 = (2w - 1) ft * w ft
To solve this equation, we can simplify it and then solve for 'w'.
10 ft^2 = (2w^2 - w) ft^2
0 = 2w^2 - w - 10
Now we have a quadratic equation, which we can solve using factoring, completing the square, or the quadratic formula. In this case, let's use factoring.
0 = (2w + 5)(w - 2)
Now we have two factors that can be set equal to zero:
2w + 5 = 0 or w - 2 = 0
Solving these equations will give us the possible values for 'w':
2w + 5 = 0
2w = -5
w = -5/2
Since we're dealing with physical dimensions, we can ignore the negative value. Therefore, 'w' cannot be -5/2.
w - 2 = 0
w = 2
So, the width of the table is 2 ft. Now we can substitute this value back into our expression for the length:
length = 2w - 1
length = 2(2) - 1
length = 3 ft
Therefore, the dimensions of the table should be 2 ft by 3 ft.