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
Answer:
Step-by-step explanation:
-10 ( y-4) + 3y = 5 <--- substitute the value of x into first equation
-10y + 40 + 3y = 5
-7y + 40 = 5
-7y = -35
y= 5
x = 5 - 4
x = 1
First integer = x
Second integer = x + 2
Third integer = x + 4
Since four times the first integer equals six more than the product of two and the third integer.
4x = 6 + 2(x + 4)
4x = 6 + 2x + 8
2x = 14
x = 7.
Hence,
First integer = x = 7
Second integer = x + 2 = 7 + 2 = 9
Third integer = x + 4 = 7 + 4 = 11.
hope this helps
Answer:
True
Step-by-step explanation:
Answer:
t = D/r
Step-by-step explanation:
you rearrange the equation so that is the subject. when you bring something over the equal sign, it reverses the function so D = r×t becomes t = D/r
