We want to construct a closed rectangular box whose base has a length three times the width. The surface area has to be 600 cm2. Determine the dimensions of the box that will produce the largest volume.
L,w,h
area = 2 Lw + 2 Lh + 2wh = 600
so
Lw+Lh+wh = 300
but L = 3w
so
3 w^2 + 4 w h = 300
now
V = L w h = 3 w^2 h
so h = V/3w^2
3 w^2 + 4 w(V/3w^2) = 300
4 V/3w = 300 - 3 w^2
(4/3)V = 300 w - 3 w^3
we want d(4/3)V /dw = 0
300 = 9 w^2
w = (10/3)sqrt 3
L = 10 sqrt 3
w L = 100
so h = 6
check my arithmetic!
Thank you!!!
You are welcome.
To determine the dimensions of the box that will produce the largest volume, we need to first express the volume of the box in terms of a single variable. Let's assume the width of the box is "w" cm.
Given that the length of the base is three times the width, the length of the base will be 3w cm. The height of the box is h cm.
Now, let's calculate the surface area of the box. The surface area of a closed rectangular box can be calculated using the formula:
Surface Area = 2lw + 2lh + 2wh
In this case, we have:
600 cm^2 = 2(3w)(w) + 2(3w)(h) + 2wh
600 cm^2 = 6w^2 + 6wh + 2wh
600 cm^2 = 6w^2 + 8wh
Next, we need to express the volume of the box in terms of a single variable. The volume of a rectangular box can be calculated by multiplying the width, length, and height. In this case, the volume (V) is given by:
V = lwh
V = (3w)(w)(h)
V = 3w^2h
Now, we can express the volume (V) in terms of a single variable by substituting the value of h from the surface area equation into the volume equation.
From the surface area equation:
600 cm^2 = 6w^2 + 8wh
8wh = 600 - 6w^2
h = (600 - 6w^2) / 8w
Substituting the expression for h in terms of w into the volume equation:
V = 3w^2h
V = 3w^2((600 - 6w^2) / 8w)
Simplifying further:
V = 3w^2(75 - 3w^2) / 8w
V = (225w - 9w^3) / 8
To find the maximum volume, we need to find the critical points of the volume equation. We can do this by taking the derivative of the volume equation with respect to w and setting it equal to zero.
dV/dw = (225 - 27w^2) / 8 = 0
225 - 27w^2 = 0
27w^2 = 225
w^2 = 225 / 27
w^2 = 25/3
w = ±√(25/3)
w ≈ ±2.886
Since length and width cannot be negative, we can discard the negative solution. So, the width of the box is approximately 2.886 cm.
To find the other dimensions, we can substitute the value of w into the equations:
Length (l) = 3w
Length (l) = 3(2.886)
Length (l) ≈ 8.658 cm
Height (h) = (600 - 6w^2) / 8w
Height (h) = (600 - 6(2.886)^2) / 8(2.886)
Height (h) ≈ 9.3 cm
Therefore, the dimensions of the box that will produce the largest volume are approximately:
Width: 2.886 cm
Length: 8.658 cm
Height: 9.3 cm