Answer :
The contractor's cost for building the walls, C, as a function of one of the dimensions of the warehouse's rectangular floor, x, is:
[tex]\(C(x) = 475x + \frac{700000}{x}\)[/tex]
To express the contractor's cost for building the walls, C, as a function of one of the dimensions of the warehouse's rectangular floor, x, we need to consider both the exterior walls and the interior wall separately and then add them up.
Let's assume that the length of the rectangular floor is x feet, and the width is y feet. We know that the area of the rectangular floor is 4000 square feet, so:
[tex]\(xy = 4000\)[/tex]
Now, we need to calculate the length of the walls that need to be built for the exterior and interior walls.
1. Exterior Walls:
The exterior walls will surround the entire warehouse, so we need to calculate the perimeter of the rectangular floor, which is the sum of all four sides.
Perimeter (P) = 2x + 2y
The cost of the exterior walls is $175 per linear foot, so the cost for the exterior walls ([tex]C_{exterior}[/tex]) is:
[tex]\(C_{\text{exterior}} = 175 \cdot P\)[/tex]
Substitute the expression for perimeter:
[tex]\(C_{\text{exterior}} = 175 \cdot (2x + 2y)\)[/tex]
2. Interior Wall:
The interior wall will run along one of the dimensions of the rectangular floor. Since it separates the warehouse into two rooms, it will have a length equal to one of the dimensions of the rectangular floor. In this case, it will have a length of x feet.
The cost of the interior wall is $125 per linear foot, so the cost for the interior wall ([tex]C_{interior}[/tex]) is:
[tex]\(C_{\text{interior}} = 125 \cdot x\)[/tex]
Now, we can express the total cost C as a function of x by adding the costs of the exterior and interior walls:
[tex]\(C(x) = C_{\text{exterior}} + C_{\text{interior}}\)[/tex]
Substitute the expressions for [tex]\(C_{\text{exterior}}\) and \(C_{\text{interior}}\):[/tex]
[tex]\(C(x) = 175 \cdot (2x + 2y) + 125 \cdot x\)[/tex]
Now, recall that [tex]\(xy = 4000\)[/tex]. You can solve for y:
[tex]\(y = \frac{4000}{x}\)[/tex]
Substitute this expression for y into the equation for C(x):
[tex]\(C(x) = 175 \cdot (2x + 2 \cdot \frac{4000}{x}) + 125 \cdot x\)[/tex]
Simplify further:
[tex]\(C(x) = 350x + \frac{700000}{x} + 125x\)[/tex]
Combine like terms:
[tex]\(C(x) = 475x + \frac{700000}{x}\)[/tex]
So, the contractor's cost for building the walls, C, as a function of one of the dimensions of the warehouse's rectangular floor, x, is:
[tex]\(C(x) = 475x + \frac{700000}{x}\)[/tex]
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The contractor's cost for building the walls as a function of one of the dimensions of the warehouse's rectangular floor are; x = 58.62 feet and y = 85.3 feet
Given that, the cost of exterior walls is $175 per linear foot.
The cost of interior walls is $100 per linear foot.
xy = 4000
y = 4000 /x
For the exterior walls, we have 2(x + y)(110)
For the interior wall, we have 100x
The cost function = C
C = 2(x + y)(125 ) + 100x
C = 320x + 240y
Recall that y = 4000 /x
C = 320x + 220(4000 /x)
C = 320x + 1100000/x
Differentiate C with respect to x
C'(x) = 320 - 1100000/x²
= (320x² -1100000) / x²
To minimize cost, C'(x) = 0
(320x² -1100000) / x² = 0
320x² -1100000 = 0
320x² = 1100000
x = 58.62 feet
Recall that y = 4000 /x
y = 4000 /58.62
y = 85.3 feet
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