Respuesta :
Answer:
the dimensions of the most economical shed are height = 10 ft and side 5 ft
Step-by-step explanation:
Given data
volume = 250 cubic feet
base costs = $4 per square foot
material for the roof costs = $6 per square foot
material for the sides costs = $2.50 per square foot
to find out
the dimensions of the most economical shed
solution
let us consider length of side x and height is h
so we can say x²h = 250
and h = 250 / x²
now cost of material = cost of base + cost top + cost 4 side
cost = x²(4) + x²(6) + 4xh (2.5)
cost = 10 x² + 10xh
put here h = 250 / x²
cost = 10 x² + 10x (250/ x² )
cost = 10 x² + (2500/ x )
differentiate and we get
c' = 20 x - 2500 / x²
put c' = 0 solve x
20 x - 2500 / x² = 0
x = 5
so we say one side is 5 ft base
and height is h = 250 / x²
h = 250 / 5²
height = 10 ft
This is about Optimization in Mathematics.
Length of base = 5 ft
Width of base = 5 ft
Height of box = 10 ft
- Let the length of the base be x
- Let the height of box be y
- We are given Volume = 250 ft³
Volume of a box is; V = length × width × height
Since the base is a square, then length = width = x
Thus;
V = x²y
x²y = 250
y = 250/x²
- The base will have the same area with the roof.
Thus, cost of base = $4x²
Cost of roof = $6x²
Area of the side is xy and so cost of 4 sides = 4(2.5xy) = 10xy
- Thus;
Total Cost; C = 4x² + 6x² + 10xy
C = 10x² + 10xy
Putting 250/x² for y, we have;
C = 10x² + 2500/x
- The dimension for the most economical cost will occur when dC/dx = 0
Thus;
dC/dx = 20x - 2500/x²
At dC/dx = 0;
20x - 2500/x² = 0
20x = 2500/x²
x³ = 2500/20
x³ = 125
x = ∛125
x = 5 ft
- Putting 5 for x in y = 250/x²;
y = 250/5²
y = 250/25
y = 10 ft
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