Answer:
the final dimensions are R= 0.409 m and L= 0.818 m
Step-by-step explanation:
Since the cost is proportional to the area , then we should find the can with volume V and minimum area. the volume is
V= π*L*R²
where L is length , R is radius
then since the area of the can = area of the sides + area of the cap and bottom .
A= 2*π*R*L + 2*π*R² , V= π*L*R² → L= V/(π*R²)
A= 2*π*R* V/(π*R²) + 2*π*R² = 2*V/R + 2*π*R²
then we find the minimum area when the derivative of the area with respect to R is 0 :
dA/dR = -2*V/R² + 4*π*R = 0 → R= ∛[V/(2π)]
replacing values
R= ∛[V/(2π)] = ∛[0.430 m³/(2π)] = 0.409 m
thus
L= V/(π*R²)= 0.430 m³/ (π*(0.409 m)²)= 0.818 m
therefore the final dimensions are
R= 0.409 m and L= 0.818 m