What is an example of a word problem
1 answer:
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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
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