A cylindrical can, open at the top, is to hold 250cm^(3) of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use pi to represent \pi . Radius = Height=

To minimize the amount of material needed to manufacture the cylindrical can, we need to find the dimensions (radius and height) that will minimize the surface area of the can. Let's denote the radius of the can as r (in cm) and the height as h (in cm). The volume of a cylinder is given by the formula V = πr^2h. Given that the volume of the can is 250 cm^3, we have: πr^2h = 250 We need to minimize the surface area of the can, which is given by the formula: A = 2πrh + πr^2 We can rewrite the equation for the volume as: h = 250 / (πr^2) Substituting this value of h into the equation for the surface area, we have: A = 2πr(250 / (πr^2)) + πr^2 = 500 / r + πr^2 To find the dimensions that minimize the surface area, we need to find the critical points of A. This can be done by taking the derivative of A with respect to r and setting it equal to zero: dA/dr = -500/r^2 + 2πr = 0 Simplifying this equation, we have: -500/r^2 + 2πr = 0 Multiplying through by r^2, we get: -500 + 2πr^3 = 0 Solving for r, we have: 2πr^3 = 500 r^3 = 500 / (2π) r^3 = 250 / π Taking the cube root of both sides, we get: r = (250 / π)^(1/3) Therefore, the radius of the can that minimizes the amount of material needed to manufacture it is (250 / π)^(1/3). To find the height, we can substitute this value of r back into the equation for h: h = 250 / (πr^2) h = 250 / (π(250 / π)^(2/3)) Simplifying this expression, we get: h = (250 / π)^(1/3) Therefore, the height of the can that minimizes the amount of material needed to manufacture it is (250 / π)^(1/3). So, the radius is (250 / π)^(1/3) and the height is also (250 / π)^(1/3).