(1 point) A rectangular storage container with an open top is to have a volume of 12 cubic meters. The length of its base is twice the width. Material for the base costs 11 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container.

The cost of the materials for the cheapest such container is approximately $249.8

Step-by-step explanation:

The volume the rectangular container is to have = 12 m³

The base length = 2 × The width of the base

The cost of the material for the base = 11 dollars/m²

The cost of the other sides = 9 dollars/m²

Let 'l', 'w', and 'h' represent the length and width of the base and the height of the rectangular storage container respectively, we have;

l = 2·w

l × w × h = 12 m³

∴ h = 12/(l × w) = 12/(2·w × w) = 6/w²

The cost of the open top rectangular container = The cost of the base + The cost of the 4 sides

The cost of the base = The area of the base × $11/m²

∴ The cost of the base = l × w × 11 =2·w × w × 11 = 22·w²

The cost of the 4 sides = The area of the four sides × $9/m²

∴ The cost of the 4 sides = 2 × (l × h + w × h) × 9

The cost of the 4 sides = 2 × (2·w × 6/w² + w × 6/w²) × 9 = 18 × (12/w + 6/w) = 324/w

∴ The cost of the open top rectangular container = 22·w² + 324/w

The coefficient of w² in the equation of the cost is positive, therefore, the cost of the materials for the cheapest such container, 'C', is given by value of 'w' at the minimum point of the equation, 22·w² + 324/w, which is given by the equating the derivative of the equation to zero as follows;

d(22·w² + 324/w)/dw = 0

∴ 44·w - 324/w² = 0

∴ 44·w = 324/w²

w³ = 324/44 = 81/11

w = ∛(81/11)

∴ C = 22·w² + 324/w = 22 × (∛(81/11))² + 324/(∛(81/11)) = 486/(∛(81/11)) ≈ 249.8

The cost of the materials for the cheapest such container, C ≈ $249.8

understand that i don't know the answers

will someone answer his question

y > 1/3x

y < = -x + 4

hope this : )