This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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To find the radius that minimizes the metal used for a closed cylindrical tin with a given capacity, we need to use calculus. Given: Capacity (Volume) of the cylinder, V = 250 ml. We assume 1 ml = 1 cm^3, so V = 250 cm^3. Let r be the radius and h be the height of the cylinder. Step 1: Write down the formulas for the volume and surface area of a closed cylinder. The volume of a cylinder is given by: V = r^2 h The surface area of a closed cylinder (area of metal used) is given by: A = 2 r^2 + 2 rh The 2 r^2 accounts for the top and bottom circular bases, and 2 rh accounts for the curved side. Step 2: Express the height (h) in terms of the radius (r) using the given volume. We are given V = 250 cm^3. 250 = r^2 h Divide both sides by : 250 = r^2 h Solve for h: h = (250)/(r^2) Step 3: Substitute the expression for h into the surface area formula to get A as a function of r only. A(r) = 2 r^2 + 2 r ((250)/(r^2)) A(r) = 2 r^2 + (500 r)/(r^2) A(r) = 2 r^2 + (500)/(r) Step 4: Differentiate the area function A(r) with respect to r. To find the minimum area, we need to find the critical points by setting the first derivative to zero. A(r) = 2 r^2 + 500 r^-1 (dA)/(dr) = (d)/(dr)(2 r^2 + 500 r^-1) (dA)/(dr) = 2 (2r) + 500 (-1 r^-2) (dA)/(dr) = 4 r - (500)/(r^2) Step 5: Set the derivative (dA)/(dr) to zero and solve for r. 4 r - (500)/(r^2) = 0 4 r = (500)/(r^2) Divide both sides by : 4r = (500)/(r^2) Multiply both sides by r^2: 4r^3 = 500 r^3 = (500)/(4) r^3 = 125 Take the cube root of both sides: r = [3]125 r = 5 cm Step 6: (Optional) Verify that this radius corresponds to a minimum area using the second derivative test. (d^2A)/(dr^2) = (d)/(dr)(4 r - 500 r^-2) (d^2A)/(dr^2) = 4 - 500 (-2 r^-3) (d^2A)/(dr^2) = 4 + (1000)/(r^3) Substitute r=5: (d^2A)/(dr^2) = 4 + (1000)/((5)^3) (d^2A)/(dr^2) = 4 + (1000)/(125) (d^2A)/(dr^2) = 4 + 8 (d^2A)/(dr^2) = 12 Since 12 > 0, the surface area is indeed a minimum at r=5 cm. The radius of the tin that minimizes the metal used is 5 cm. That's 2 down. 3 left today — send the next one.