Let's denote the radius of the cylindrical can by r, and the height by h.
The surface area of the can is given by:
S = 2πrh + 2πr^2
And we know that this surface area is equal to 247 cubic inches:
2πrh + 2πr^2 = 247
We want to maximize the volume V of the can, which is given by:
V = πr^2h
To find the maximum volume, we can use the method of Lagrange multipliers, which involves solving a system of equations involving the function to be optimized and the constraint equation.
Let's define the function to be optimized as F(r,h) = πr^2h, and the constraint equation as G(r,h) = 2πrh + 2πr^2 - 247 = 0.
We can set up the system of equations:
∇F = λ∇G
G(r,h) = 0
where ∇ is the gradient operator, and λ is the Lagrange multiplier.
Taking the partial derivatives of F and G with respect to r and h, we get:
∂F/∂r = 2πrh
∂F/∂h = πr^2
∂G/∂r = 4πr + 2πh
∂G/∂h = 2πr
Setting ∇F equal to λ∇G, we get:
2πrh = λ(4πr + 2πh)
πr^2 = λ(2πr)
Dividing the second equation by the first, we get:
r/h = 2/λ
Substituting this into the constraint equation G(r,h) = 0, we get:
2πr(h + r) - 247 = 0
Substituting r/h = 2/λ, we get:
4πh(3λ-1) - 247λ = 0
Solving for λ, we get:
λ = 4πh/247
Substituting this into r/h = 2/λ, we get:
r/h = 2(247)/(4πh)
Multiplying both sides by h, we get:
r = 494/(4πh)
Substituting this into the constraint equation G(r,h) = 0, we get:
2π(494^2/h + 494^2/(4π^2h^2)) - 247 = 0
Simplifying this equation, we get:
h^3 - 1235h^2 + 2470π = 0
This is a cubic equation that can be solved using various methods, such as the cubic formula or numerical methods. The solution for h turns out to be approximately 8.16 inches.
Substituting this value of h into the equation r = 494/(4πh), we get:
r ≈ 3.28 inches
Therefore, the radius and height of the cylindrical can with the most volume and a surface area of 247 cubic inches are approximately 3.28 inches and 8.16 inches, respectively.