5% ( interest income from a muni is exempt from federal income tax so the after-tax yield equals the before tax yield)
Answer:
The dimensions are 1 m × 1 m × 5 m
Explanation:
Let the dimension be x, y, z
Volume = xyz = 5 m³ ................(1)
According to question:
Cost function, C = 110(xy) + 30(xy) + 14(2xz+2yz)
or
C = 140xy + 28xz + 28yz ..........(2)
We need to maximise (2)
Given condition (1)
Using the concept of lagranges multipliers
...........(3)
...........(4)
...........(5)
From (3) and (4) and(5)
140xy + 28xz = 140xy + 28yz = 28xz + 28yz
thus,
140xy + 28xz = 140xy + 28yz
or
28xz = 28 yz
or
x = y ............(a)
140xy + 28yz = 28xz + 28yz
substituting x from (a)
140(y)y + 28yz = 28(y)z + 28yz
or
140y² = 28yz
or
5y = z
and,
volume = 5 m³
or
xyz = 5 m³
or
x(x)(5y) = 5 m³
or
x²(5x) = 5 m³
or
5x³ = 5 m³
or
x = 1 m
Hence,
y = x = 1 m
and,
z = 5y = 5(1) = 5 m
Therefore,
The dimensions are 1 m × 1 m × 5 m
Answer:
total cost of farming = $380
so correct option is d. $380
Explanation:
given data
cost of seeds = $130
Farmer Ziva charges = $25
time = 10 hours
solution
so total cost of farming is calculated as
total cost of farming = cost of seeds + opportunity cost
so put value
total cost of farming = $130 + ( $25 × 10 )
total cost of farming = 130 + ( 250 )
total cost of farming = $380
so correct option is d. $380
Answer:
a, 22276.07
b. $32.9157 million
c.$29.9669million
Explanation:
Find the values of k and a assuming a relationship of the form Assume that f(y)=ky^a is in units of barrels per day.
b. Determine the optimal timing of plant additions and the optimal size and cost of each plant addition.a=0.8073, rx=0.41
optimal timing x=rx/r=2.05yrs
optimal size xD=2.05(1.5)
3.075million barrels/year
$32.9157 million
c. Suppose that the largest single refinery that can be built with current technology is 7,500 barrels per day. Determine the optimal timing of plant additions and the optimal size and cost of each plant in this case
Optimal size xD=min
Optimal timing will be X^*=x*D/D=2.7375/1.5=1.825 year
optimal cost f(y)=ky^a=0.0223(7500)^0.8073=$29,9669 milion
Answer:
In order to evaluate which ordering quantity is least costly, you should calculate the total cost associated with the range 1 to 99 units, using a quantity of___75___ units, and the range 200 units and up, using a quantity of____190____ units.
Explanation:
a) Data:
Quantity Price per unit Calculated EOQ
1 to 99 $110 75
100 to 199 $80 120
200 and up $50 190
b) The calculated Economic Order Quantity (EOQ) is the optimal quantity that minimizes inventory costs which include holding, shortage, and ordering or production setup costs. The EOQ is computed as the square root of: [2(setup costs)(demand rate)] / holding costs. While EOQ is popular, its use is criticized on the ground that it is over-simplistic and relies largely on consistent data inputs. Consistent data does not reflect reality.