Respuesta :
This is a calculus problem (in optimization). Let the aquarium dimensions be L, W and H. Then L*W*H = 256 ft^3. But W=L here, so W^2*H = 256 ft^3.
Write an expression for the area of the aquarium's glass sides and bottom.
256 ft^3 256 ft^3
A= W^2 + 4(W*H) = W^2 + 4*W*(------------- ) = W^2 + 4*-------------------
W^2 W
So now we have A(W), the area as a function of W alone.
We want to minimize this area. To do this, differentiate A(W) with respect to W and set the result = to 0. We want to determine the "critical numbers."
dA/dW = 2W - 1024*W^(-2) = 0
1024
Then 2W = ---------------
W^2
2W^3 = 1024, so W^3 = 512, and W = third root of 512 = 8
If W = 8 ft, then L = 8 ft also. Since L*W*H = 256 ft^3,
L*W*H = 256 ft^3 = (8 ft)^2*H = 256 ft^3. Then H = 4
The acquarium dimensions are 8 by 8 by 4 feet. This keeps the area of the aquarium to a minimum.
Write an expression for the area of the aquarium's glass sides and bottom.
256 ft^3 256 ft^3
A= W^2 + 4(W*H) = W^2 + 4*W*(------------- ) = W^2 + 4*-------------------
W^2 W
So now we have A(W), the area as a function of W alone.
We want to minimize this area. To do this, differentiate A(W) with respect to W and set the result = to 0. We want to determine the "critical numbers."
dA/dW = 2W - 1024*W^(-2) = 0
1024
Then 2W = ---------------
W^2
2W^3 = 1024, so W^3 = 512, and W = third root of 512 = 8
If W = 8 ft, then L = 8 ft also. Since L*W*H = 256 ft^3,
L*W*H = 256 ft^3 = (8 ft)^2*H = 256 ft^3. Then H = 4
The acquarium dimensions are 8 by 8 by 4 feet. This keeps the area of the aquarium to a minimum.
To solve the problem we must know about the volume and the surface area of cuboids.
The length of the aquarium is 4, and the width of the aquarium is 8.
Given to us
- The aquarium must hold 256 ft^3 of water.
We know that the base of the aquarium should be a square, therefore, the length and the breadth of the aquarium will be the same.
Volume of the Aquarium
Length x breadth x height = 256 ft³
Length x Length x height = 256 ft³
Length ² x height = 256 ft³
[tex]L^2\times H = 256\\\\H = \dfrac{256}{L^2}[/tex]
Surface Area of the Aquarium
Surface Area of the Aquarium
= Area of the base + (4 x Area of the 4 sides)
[tex]= L^2 + 4(L\times H)[/tex]
Substituting the value of H,
[tex]= L^2 + 4L(\dfrac{256}{L^2})[/tex]
Length of the Aquarium
Let the surface area be a function of L, therefore,
[tex]f(L) = L^2 + 4L(\dfrac{256}{L^2})[/tex]
Differentiating the function to get the minimum,
[tex]f(L) = L^2 + 4L(\dfrac{256}{L^2})\\\\f(L) = L^2 + (\dfrac{1024}{L})\\\\f'(L) = 2L - (\dfrac{1024}{L^2})[/tex]
Equating against zero,
[tex]2L - (\dfrac{1024}{L^2}) = 0\\\\\dfrac{2L^3-1024}{L^2} = 0\\\\2L^3-1024 = 0\\\\L = 8[/tex]
Substituting the value of L,
[tex]H = \dfrac{256}{L^2}\\\\H = \dfrac{256}{8^2}\\\\H = 4[/tex]
But this way the area will be more, therefore, replace the value of L with 8,
L = 4
H = 8
Hence, the length of the aquarium is 4, and the width of the aquarium is 8.
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