The area of a rectangle is $63 \mathrm{yd}^{2}$, and the length of the rectangle is $11 \mathrm{yd}$ more than twice the width. Find the dimensions of the rectangle.

Step 1 ：Let's denote the width of the rectangle as \(w\) and the length as \(l\). We know that the area \(A\) of a rectangle is given by the formula \(A = l \times w\). We are given that \(A = 63 \, \text{yd}^{2}\) and \(l = 2w + 11 \, \text{yd}\).

Step 2 ：We can substitute the equation for \(l\) into the equation for \(A\) to get a quadratic equation in terms of \(w\): \(w(2w + 11) = 63\).

Step 3 ：Solving this quadratic equation, we find two solutions for \(w\): -9 and \(\frac{7}{2}\). However, the width of a rectangle cannot be negative, so we discard the solution -9.

Step 4 ：Therefore, the width of the rectangle is \(\frac{7}{2}\) yards. Substituting this value into the equation for \(l\), we find that the length of the rectangle is 18 yards.

Step 5 ：Final Answer: The dimensions of the rectangle are \(\boxed{\frac{7}{2} \, \text{yd}}\) by \(\boxed{18 \, \text{yd}}\).