Answer:
the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm
Step-by-step explanation:
since the volume of a cylinder is
V= π*R²*L → L =V/ (π*R²)
the cost function is
Cost = cost of side material * side area + cost of top and bottom material * top and bottom area
C = a* 2*π*R*L + b* 2*π*R²
replacing the value of L
C = a* 2*π*R* V/ (π*R²) + b* 2*π*R² = a* 2*V/R + b* 2*π*R²
then the optimal radius for minimum cost can be found when the derivative of the cost with respect to the radius equals 0 , then
dC/dR = -2*a*V/R² + 4*π*b*R = 0
4*π*b*R = 2*a*V/R²
R³ = a*V/(2*π*b)
R= ∛( a*V/(2*π*b))
replacing values
R= ∛( a*V/(2*π*b)) = ∛(0.03$/cm² * 600 cm³ /(2*π* 0.05$/cm²) )= 3.85 cm
then
L =V/ (π*R²) = 600 cm³/(π*(3.85 cm)²) = 12.88 cm
therefore the dimensions that minimize the cost of the cylinder are R= 3.85 cm and L=12.88 cm
so,it is the answer that you are looking for.please mark me as brainliest.
Answer:X2 = 5
Y2 = 4
ΔX = 4
ΔY = 3
θ = 36.869897645844°
Equation of the line:
y = 0.75x + 0.25
When x=0, y = 0.25
When y=0, x = -0.33333333333333
OR
X2 = -3
Y2 = -2
ΔX = -4
ΔY = -3
θ = 216.86989764584°
Equation of the line:
y = 0.75x + 0.25
When x=0, y = 0.25
When y=0, x = -0.33333333333333
After simplifying, the answer comes to x-y. I Hope this helped :).
