V=At A confectioner is applying a sugar coating to some spherical candies. The given equation approximates V, the volume of coating in cubic millimeters, needed to create a t millimeter thick coating on a candy with a surface area of A square millimeters. If 20 pi cubic millimeters of coating is needed to create a 0.2 -millimeter thick coat on a spherical candy, what is the uncoated candy's radius in millimeters? (The formula for the surface area, A, of a sphere with radius r is A=4pir^(2).)

Question

V=At A confectioner is applying a sugar coating to some spherical candies. The given equation approximates  V, the volume of coating in cubic millimeters, needed to create a  t millimeter thick coating on a candy with a surface area of  A square millimeters. If  20 pi cubic millimeters of coating is needed to create a 0.2 -millimeter thick coat on a spherical candy, what is the uncoated candy's radius in millimeters? (The formula for the surface area,  A, of a sphere with radius  r is  A=4pir^(2).)

Bytelearn · Accepted Answer

Given Volume and Thickness: We are given the volume of the coating V as 20 pi cubic millimeters and the thickness of the coating t as 0.2 millimeters. We need to find the surface area A of the uncoated candy to eventually find its radius r. Using the given equation V = At, we can solve for A. V = 20 pi t = 0.2 A = V / tCalculate Surface Area: Now we calculate A using the values of V and t. A = (20 pi) / 0.2 A = 100 pi square millimetersUse Surface Area Formula: The formula for the surface area A of a sphere with radius r is A = 4 pi r^2. We can set this equal to the calculated surface area and solve for r. 100 pi = 4 pi r^2Isolate and Solve for r: Divide both sides of the equation by 4 pi to isolate r^2. (100 pi) / (4 pi) = r^2 25 = r^2Final Radius Calculation: Take the square root of both sides to solve for r. r = sqrt(25) r = 5 millimeters

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