4 attempts remaining.

Find the length $L$ and width $W$ (with $W \leq L$ ) of the rectangle with area 361 square inches that has minimum perimeter, and then find the minimum perimeter.
\[
\begin{array}{l}
L= \\
W=
\end{array}
\]

Minimum perimeter $=$
Submit answer

Step 1 ：Let's denote the length of the rectangle as \(L\) and the width as \(W\). The area of the rectangle is given by the product of its length and width, so we have \(L \times W = 361\).

Step 2 ：The perimeter of a rectangle is given by \(2L + 2W\). We want to minimize this quantity.

Step 3 ：Express \(W\) in terms of \(L\) using the area equation: \(W = 361 / L\).

Step 4 ：Now we can express the perimeter in terms of \(L\) only: \(P = 2L + 2(361 / L)\).

Step 5 ：To find the minimum of this function, we can take the derivative and set it equal to zero. The derivative of \(P\) with respect to \(L\) is: \(dP/dL = 2 - 722 / L^2\).

Step 6 ：Setting this equal to zero gives: \(2 - 722 / L^2 = 0\).

Step 7 ：Solving for \(L\) gives: \(L^2 = 722 / 2\).

Step 8 ：\(L = \sqrt{361} = 19\).

Step 9 ：Substituting \(L = 19\) into the equation \(W = 361 / L\) gives: \(W = 361 / 19 = 19\).

Step 10 ：So the rectangle is actually a square with side length 19 inches.

Step 11 ：The minimum perimeter is then \(2L + 2W = 2*19 + 2*19 = 76\) inches.

Step 12 ：So, the length \(L = 19\) inches, the width \(W = 19\) inches, and the minimum perimeter is 76 inches.

Step 13 ：The area of the rectangle is indeed \(19*19 = 361\) square inches.

Step 14 ：The perimeter of the rectangle is indeed \(2*19 + 2*19 = 76\) inches.

Step 15 ：The rectangle with given area 361 square inches that has minimum perimeter is a square, which is consistent with the properties of rectangles.

Step 16 ：\(\boxed{\text{So, the length } L = 19 \text{ inches, the width } W = 19 \text{ inches, and the minimum perimeter is 76 inches.}}\)