A V-shaped water gutter is to be considered rectangular sheets of metal 10 inches wide. Find the angle between the sheets that will maximize the carrying capacity of the gutter.

The carrying capacity of the gutter depends on the cross-sectional area of water it can hold. If we assume that the depth of the gutter is constant along its length, then the cross-sectional area is proportional to the square of the width of the gutter. Let's call the angle between the sheets "theta". Then, the width of the gutter at the top can be expressed as 10 cos(theta), while the width at the bottom (assuming straight sides) is 20 sin(theta). Therefore, the cross-sectional area of the gutter is: A = (1/2) * (10 cos(theta) + 20 sin(theta))^2 * h where h is the constant depth of the gutter. To find the maximum value of A, we can take the derivative with respect to\theta and set it equal to zero: dA/dtheta = 20h(sin(theta) - cos(theta))(5 sin(theta) + 2 cos(theta)) = 0 This equation has two solutions:\theta = arctan(5/2) and\theta = arctan(-2/5). Since the angle between the sheets cannot be negative, we take the positive solution: theta = arctan(5/2) ? 68.2 degrees Therefore, the angle between the sheets that maximizes the carrying capacity of the V-shaped water gutter is about 68.2 degrees.