Refer to the diagram shown below.
The volume of the container is 10 m³, therefore
x*2x*h = 10
2x²h = 10
h = 5/x² (1)
The base area is 2x² m².
The cost is $10 per m², therefore the cost of the base is
(2x²)*($10) = 20x²
The area of the sides is
2hx + 2(2xh) = 6hx = 6x*(5/x²) = 30/x m²
The cost is $6 per m², therefore the cost of the sides is
(30/x)*($6) = 180/x
The total cost is
C = 20x² + 180/x
The minimum cost is determined by C' = 0.
That is,
40x - 180/x² = 0
x³ = 180/40 = 4.5
x = 1.651
The second derivative of C is
C'' = 40 + 360/x³
C''(1.651) = 120 >0, so x = 1.651 m yields the minimum cost.
The total cost is
C = 20(1.651)² + 180/1.651 = $163.54
Answer: $163.54
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Answer:
4. dy/dx = -2
8. dy/dx = 1/2 x^(-3/2)
10/ dy/dr = 4 pi r^2
Step-by-step explanation:
4. y = -2x+7
dy/dx = -2(1)
dy/dx = -2
8. y = 4 - x^-1/2
dy/dx = - (-1/2x^ (-1/2-1)
dy/dx = 1/2 x^(-3/2)
10. y = 4/3 pi r^3
dy/dr = 4/3 pi (3r^2)
dy/dr = 4 pi r^2
Answer:
17.907082 unit
Step-by-step explanation:
According to the Question,
Given, A circle with centre F, ∠EFG=54 and EF=19 .
length of arc EG = Radius(EF) × ∠EFG(in Radian)
- We Know, 1 degree = 0.0174533 Radian
- 54 degree = 0.942478 Radian
length of arc EG = 19 x 0.942478 ⇔ 17.907082 unit
(For Diagram please find in attachment)
I think the answer to y when x=4 and z=15; if y varies jointly as x and z y=5 when z=8 and x=10 is B. Y=1/2. Y•15; 12/4.