Respuesta :
Answer:
When dimension of box is 33.33 inches × 33.33 inches ×8.33 then its volume is maximum and is 9259.26 cubic inches.
Step-by-step explanation:
Let h be the length (in inches) of the square corners that has been cut out from the cardboard and that would be the height of the cardboard box.
Since the squares have been cut from cardboard, both sides of the cardboard would reduce by 2h.
Thus, The dimension of box is (50 – 2h) × (50 – 2h) × h in dimensions.
The volume V of rectangular box = (Length × Breadth × Height) cubic inches.
[tex]V=(50-2h) \times (50-2h) \times h[/tex]
[tex]V=(50-2h)^2 \times h[/tex] ..............(1)
Using [tex](a-b)^2=a^2+b^2-2ab[/tex]
[tex]V=h(2500+4h^2-200h)[/tex]
[tex]V=2500h+4h^3-200h^2[/tex]
For obtaining a box of maximum volume, maximize V as a function of h.
Differentiate both sides with respect to h,
[tex]\frac{dV}{dh}=2500+12h^2-400h[/tex]
[tex]\frac{dV}{dh}=4(625+3h^2-100h)[/tex]
Solving quadratic equation,[tex]625+3h^2-100h[/tex]
[tex]\frac{dV}{dh}=4(3h^2-25h-75h+625)[/tex]
[tex]\frac{dV}{dh}=4(h(3h-25)-25(3h-25))[/tex]
[tex]\frac{dV}{dh}=4((h-25)(3h-25))[/tex]
For maximum, [tex]\frac{dV}{dh}=0[/tex]
thus,[tex]4((h-25)(3h-25))=0[/tex]
⇒ [tex]h= 25[/tex] or [tex]h=\frac{25}{3}[/tex]
Now check (1) for [tex]h= 25[/tex] and [tex]h=\frac{25}{3}[/tex].
[tex]h= 25[/tex] is not possible as when h is 25 inches then length and breadth becomes 0.
When [tex]h=\frac{25}{3}[/tex].
(1) ⇒ [tex]V=(50-2(\frac{25}{3}))^2 \times\frac{25}{3}=9259.2592593[/tex]
This is the maximum volume the box can assume.
Thus, when dimension of box is 33.3 inches × 33.3 inches ×8.3 then its volume is maximum and is 9259.26 cubic inches.
The dimension that maximizes the volume is 33.4 by 33.4 by 8.3 inches, while the maximum volume is 9259.1 cubic inches
The dimension of the cardboard is given as:
[tex]\mathbf{Length = Width = 50}[/tex]
Assume the cut-out is x.
So, the dimension of the box is:
[tex]\mathbf{Length = Width = 50 -2x}[/tex]
[tex]\mathbf{Height = x}[/tex]
The volume is then calculated as:
[tex]\mathbf{Volume = Length \times Width \times Height}[/tex]
This gives
[tex]\mathbf{Volume = (50 -2x) \times (50 -2x) \times x}[/tex]
Expand
[tex]\mathbf{Volume = 2500x - 200x^2 + 4x^3}[/tex]
Differentiate
[tex]\mathbf{V' = 2500 - 400x + 12x^2}[/tex]
Set to 0
[tex]\mathbf{2500 - 400x + 12x^2 = 0}[/tex]
Using a calculator, we have:
[tex]\mathbf{x = 25, 8.3}[/tex]
x = 25 is greater than the dimension of the box.
So, we have:
[tex]\mathbf{x = 8.3}[/tex]
Substitute [tex]\mathbf{x = 8.3}[/tex] in [tex]\mathbf{Length = Width = 50 -2x}[/tex]
[tex]\mathbf{Length = Width = 50 -2 \times 8.3}[/tex]
[tex]\mathbf{Length = Width = 33.4}[/tex]
Recall that: [tex]\mathbf{Volume = Length \times Width \times Height}[/tex]
So, we have:
[tex]\mathbf{Volume = 33.4 \times 33.4 \times 8.3}[/tex]
[tex]\mathbf{Volume = 9259.1}[/tex]
Hence, the maximum volume is 9259.1 cubic inches
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