Answer:
Side length = [tex]\sqrt{\frac{A}{3} }[/tex] cm , Height = [tex]\frac{1}{2} \sqrt{\frac{A}{3} }[/tex] cm , Volume = [tex]\frac{A\sqrt{A}}{6\sqrt{3} }[/tex] cm³
Step-by-step explanation:
Assume
Side length of base = x
Height of box = y
total material required to construct box = A ( given in question)
So it can be written as
A = x² + 4xy
4xy = A - x²
Volume of box = Area x height
V = x² ₓ y
V = x² ₓ ( [tex]\frac{A - x^{2} }{4x}[/tex] )
V = [tex]\frac{Ax - x^{3} }{4}[/tex]
To find max volume put V' = 0
So taking derivative equation becomes
[tex]\frac{A - 3 x^{2} }{4} = 0[/tex]
A = 3 [tex]x^{2}[/tex]
[tex]x^{2}[/tex] = [tex]\frac{A}{3}[/tex]
x = [tex]\sqrt{\frac{A}{3\\} }[/tex]
put value of x in equation 1
y = [tex]\frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }[/tex]
y = [tex]\frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }[/tex]
y = [tex]\frac{1}{2} \sqrt{\frac{A}{3} }[/tex]
So the volume will be
V = [tex]x^{2}[/tex] × y
Put values of x and y from equation 2 & 3
V = [tex]\frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )[/tex]
V = [tex]\frac{A\sqrt{A}}{6\sqrt{3} }[/tex]
The side length is [tex]\mathbf{l =\sqrt{ \frac A3}}[/tex], the height is [tex]\mathbf{h = \frac{1}{2}\sqrt{\frac A3}}[/tex] and the maximal volume is [tex]\mathbf{V = \frac A6 \sqrt{\frac A3}}[/tex]
Let the dimensions of the box be l and h, where l represents the base length and h represents the height.
The volume is calculated as:
[tex]\mathbf{V = l^2h}[/tex]
The surface area is:
[tex]\mathbf{A= l^2 + 4lh}[/tex]
Make h the subject
[tex]\mathbf{h = \frac{A- l^2}{4l}}[/tex]
Substitute [tex]\mathbf{h = \frac{A- l^2}{4l}}[/tex] in [tex]\mathbf{V = l^2h}[/tex]
[tex]\mathbf{V = l^2 \times \frac{A- l^2}{4l}}[/tex]
[tex]\mathbf{V = l \times \frac{A- l^2}{4}}[/tex]
[tex]\mathbf{V = \frac{Al- l^3}{4}}[/tex]
Split
[tex]\mathbf{V = \frac{Al}{4}- \frac{l^3}{4}}[/tex]
Differentiate
[tex]\mathbf{V' = \frac{A}{4}- \frac{3l^2}{4}}[/tex]
Set to 0
[tex]\mathbf{\frac{A}{4}- \frac{3l^2}{4} = 0}[/tex]
Multiply through by 4
[tex]\mathbf{A- 3l^2 = 0}[/tex]
Add 3l^2 to both sides
[tex]\mathbf{3l^2 = A}[/tex]
Divide both sides by 3
[tex]\mathbf{l^2 = \frac A3}[/tex]
Take square roots
[tex]\mathbf{l =\sqrt{ \frac A3}}[/tex]
Recall that: [tex]\mathbf{h = \frac{A- l^2}{4l}}[/tex]
So, we have:
[tex]\mathbf{h = \frac{A - \frac{A}{3}}{4\sqrt{A/3}}}[/tex]
[tex]\mathbf{h = \frac{\frac{2A}{3}}{4\sqrt{A/3}}}[/tex]
Divide
[tex]\mathbf{h = \frac{2\sqrt{A/3}}{4}}[/tex]
[tex]\mathbf{h = \frac{\sqrt{A/3}}{2}}[/tex]
Rewrite as:
[tex]\mathbf{h = \frac{1}{2}\sqrt{\frac A3}}[/tex]
Recall that:
[tex]\mathbf{V = l^2h}[/tex]
So, we have:
[tex]\mathbf{V = \frac A3 \times \frac{1}{2}\sqrt{\frac A3}}[/tex]
[tex]\mathbf{V = \frac A6 \sqrt{\frac A3}}[/tex]
Hence, the side length is [tex]\mathbf{l =\sqrt{ \frac A3}}[/tex], the height is [tex]\mathbf{h = \frac{1}{2}\sqrt{\frac A3}}[/tex] and the maximal volume is [tex]\mathbf{V = \frac A6 \sqrt{\frac A3}}[/tex]
Read more about volumes at:
https://brainly.com/question/11599887
