Answer:
Maximum volume of the box is 3042.91 cm³
Step-by-step explanation:
Area of the material available to make a box with square base and open top = 1000 square cm.
Total surface area of the box = 2(lh + bh) + l×b
Where l = length of the box
b = width of the box
h = height of the box
If l = b, Total surface area of the box = 2(lh + lh) + l×l
Surface area = 4lh + l² = 1000
l(4h + l) = 1000
4h + l = [tex]\frac{1000}{l}[/tex]
h = [tex]\frac{1000}{4l}-\frac{l}{4}[/tex]
Now volume of the square box = lbh
V = l²h
V = l²([tex]\frac{250}{l}-\frac{l}{4}[/tex])
V = 250l - [tex]\frac{l^{3}}{4}[/tex]
Now to calculate the maximum volume we have to find the derivative of the volume.
[tex]\frac{dV}{dl}=250-\frac{3l^{2} }{4}[/tex]
By equating derivative to zero,
250 - [tex]\frac{3l^{2}}{4}[/tex] = 0
l = [tex]\sqrt{\frac{250\times 4}{3}}[/tex]
l = 18.26 cm
For l = 18.26, V = 250(18.26) - [tex]\frac{(18.26)^{3}}{4}[/tex]
V = 4565 - 1522.09
= 3042.91 cm³
Therefore, maximum volume of the box is 3042.91 cm³
The largest possible volume of the box is given by the maximum value of
the function for the volume of the box.
The largest possible volume of the box is approximately 3,042.9 cm³.
Reasons:
Known parameters:
Area of the box, A = 1000 cm²
Shape of the base of the box = Square
Required:
Largest possible volume of the box
Solution:
Let y represent the side length of the square base and let x represent the
height of the box, we have;
Volume, V = y²·x
Given area, A = y² + 2·y·x + 2·y·x = y² + 4·y·x
∴ 1000 = y² + 4·y·x
Which gives;
[tex]x =\dfrac{1000 - y^2}{4 \cdot y}[/tex]
Plugging in the value of x in the volume, V formula, we get;
[tex]V =\dfrac{y^2 \cdot \left (1000 - y^2\right)}{4 \cdot y}[/tex]
Differentiating and equating the result to zero gives;
[tex]\dfrac{dV}{dy} = \dfrac{d}{dy} \left( \dfrac{y^2 \cdot \left (1000 - y^2\right)}{4 \cdot y} \right) =250 - \dfrac{3}{4} \cdot y^2 = 0[/tex]
[tex]y = \sqrt{\dfrac{1000}{3} } = \dfrac{10\cdot \sqrt{30} }{3}[/tex]
Which gives;
[tex]x = \dfrac{1000 - \left( \dfrac{10\cdot \sqrt{30} }{3} \right)^2}{4 \times \left( \dfrac{10\cdot \sqrt{30} }{3} \right)}= \dfrac{5 \cdot \sqrt{30} }{3}[/tex]
Therefore;
[tex]V = \left( \dfrac{10\cdot \sqrt{30} }{3} \right)^2 \times \dfrac{5 \cdot \sqrt{30} }{3} = \dfrac{5000 \times \sqrt{30} }{9} \approx 3,042.9[/tex]
The largest possible volume of the box is V ≈ 3,042.9 cm³.
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