To solve this problem, we will define the width and length of the rectangle in terms of a variable and then use the information provided to set up and solve an equation.

Step-by-Step Solution:

1. Define the variables:

   • Let $w$ be the width of the rectangle.
   • According to the problem, the length $l$ is "1 more than twice the width". So, we can write this as: $l = 2w + 1$

2. Set up the equation for the area:

   • The area $A$ of a rectangle is given by the product of its length and width: $A = l \times w$
   • We know the area is $28y^2$. Substitute the expressions for $l$ and $A$ into the equation: $28y^2 = (2w + 1) \times w$

3. Expand and simplify the equation:

   • Expand the equation: $28y^2 = 2w^2 + w$
   • Rearrange the equation to set it equal to 0: $2w^2 + w - 28y^2 = 0$

4. Solve the quadratic equation for $w$:

   • To solve for $w$, use the quadratic formula $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = 1$, and $c = -28y^2$.
   • Plugging in the values: $w = \frac{-1 \pm \sqrt{1^2 - 4 \times 2 \times (-28y^2)}}{2 \times 2}$
   • Simplify under the square root: $w = \frac{-1 \pm \sqrt{1 + 224y^2}}{4}$

5. Determine the width $w$:

   • Since the width must be a positive number, we take the positive root: $w = \frac{-1 + \sqrt{1 + 224y^2}}{4}$

6. Calculate the length $l$:

   • Now, substitute $w$ back into the expression for $l$: $l = 2w + 1 = 2 \left( \frac{-1 + \sqrt{1 + 224y^2}}{4} \right) + 1$
   • Simplify this expression: $l = \frac{-2 + 2\sqrt{1 + 224y^2}}{4} + 1 = \frac{-2 + 2\sqrt{1 + 224y^2} + 4}{4}$ $l = \frac{2 + 2\sqrt{1 + 224y^2}}{4} = \frac{1 + \sqrt{1 + 224y^2}}{2}$

Final Dimensions:

• The width of the rectangle is: $w = \frac{-1 + \sqrt{1 + 224y^2}}{4}$

• The length of the rectangle is: $l = \frac{1 + \sqrt{1 + 224y^2}}{2}$

Would you like more details or have any questions about this solution?

Related Questions:

1. How would the dimensions change if the area was a different expression?
2. Can you find the width and length of the rectangle if $y = 1$?
3. What happens to the dimensions if the length was defined differently, such as being 3 times the width?
4. How can we graph the relationship between width and length for different values of $y$?
5. What are the properties of the quadratic formula that help us solve this problem?

Tip:

To solve a quadratic equation, always make sure to rearrange it in the standard form $ax^2 + bx + c = 0$ and then use the quadratic formula or factorization for solutions.