The dimensions of a rectangle are such that its length is 3 in. more than its width. If the length were doubled and if the width were decreased by 1in., the area would be increased by 150 in.^2. What are the length and width of the rectangle?
Let's assume the width of the rectangle is "w" inches.
According to the given information, the length of the rectangle is 3 inches more than the width. So, the length can be represented as "w + 3" inches.
The area of a rectangle is calculated by multiplying its length and width. The initial area of the rectangle can be calculated as:
Initial area = width * length
Initial area = w * (w + 3)
If we double the length and decrease the width by 1 inch, the new dimensions of the rectangle would be (2 * (w + 3)) and (w - 1), respectively.
The new area of the rectangle can be calculated as:
New area = new width * new length
New area = (w - 1) * (2 * (w + 3))
According to the problem, the new area is 150 square inches greater than the initial area. So, we can set up the equation:
New area = Initial area + 150
Substituting the values, we get:
(w - 1) * (2 * (w + 3)) = w * (w + 3) + 150
Simplifying the equation, we get:
2w^2 + 3w - 2 = w^2 + 3w + 150
Combining like terms, we get:
w^2 = 152
Taking the square root of both sides, we get:
w = ±√152
Since the width cannot be negative, we ignore the negative value. Therefore, the width of the rectangle is √152 inches.
Substituting this value back into the initial equation for the length, we get:
Length = Width + 3
Length = √152 + 3
Calculating the values, we get:
Length ≈ 12.33 inches
Width ≈ 12.33 - 3
Width ≈ 9.33 inches
Therefore, the length and width of the rectangle are approximately 12.33 inches and 9.33 inches, respectively.
To solve this problem, let's first assign variables to represent the length and width of the rectangle.
Let's say the width of the rectangle is "w" inches. Since the length is 3 inches more than the width, its length would be represented as "w + 3" inches.
Now, let's apply the given information to form an equation:
Area of the original rectangle = length × width
Area1 = (w + 3)w
According to the problem, when the length is doubled and the width is decreased by 1 inch, the area is increased by 150 square inches. So, the new area is:
Area2 = (2(w + 3))(w - 1)
Now we can set up an equation using the given information:
Area2 - Area1 = 150
(2(w + 3))(w - 1) - (w + 3)w = 150
Simplifying this equation will allow us to find the value of "w" and then calculate the length of the rectangle.
Let's expand and simplify the equation step by step:
2(w + 3)(w - 1) - (w + 3)w = 150
2(w^2 + 3w - w - 3) - (w^2 + 3w) = 150
2(w^2 + 2w - 3) - (w^2 + 3w) = 150
2w^2 + 4w - 6 - w^2 - 3w = 150
w^2 + w - 156 = 0
Now, we have a quadratic equation. We can solve this equation by factoring, completing the square, or using the quadratic formula.
The equation factors to:
(w + 13)(w - 12) = 0
Setting each factor equal to zero gives us two possible values for "w":
w + 13 = 0 or w - 12 = 0
w = -13 or w = 12
Since the width of a rectangle cannot be negative, we discard the value of -13.
Therefore, the width of the rectangle is 12 inches.
To find the length, we can substitute this value back into the equation: length = width + 3
length = 12 + 3 = 15 inches
So, the length of the rectangle is 15 inches and the width is 12 inches.