The correct piece of aluminum with correct length and width in cm., that
makes the volume of the cylinder maximum.
- B. Length = 7 cm. width 3.5 cm.
Reasons:
When the perimeter of the rectangle is 21 cm., we have;
2·l + 2·w = 21 cm
Where, represents the length of the rectangle and w represents the width of the rectangle.
Which gives;
[tex]\displaystyle l = \frac{21 - 2\cdot w}{2} = \mathbf{10.5 -w}[/tex]
The volume of the cylinder = π·r²·w
Where;
2·π·r = l
[tex]\displaystyle r = \frac{l}{2 \cdot \pi} = \mathbf{\frac{10.5 - w}{2 \cdot \pi}}[/tex]
Which gives;
[tex]\displaystyle V = \pi \cdot r^2 \cdot w = \pi \times \left(\frac{10.5 - w}{2 \cdot \pi}\right)^2 \times w = \mathbf{ \frac{w^3-21 \cdot w^2 + 110.25 \cdot w}{4 \cdot \pi}}[/tex]
At maximum volume, we have;
[tex]\displaystyle V' = \frac{d}{dw} \left( \displaystyle \frac{w^3-21 \cdot w^2 + 110.25 \cdot w}{4 \cdot \pi} \right) = \mathbf{\frac{0.75 \cdot w^2- 10.5 \cdot w + 27.5625}{\pi}} = 0[/tex]
Which, by using a graphing calculator, gives;
(w - 3.5)·(w - 10.5)·0.75 = 0
w = 3.5 or 10.5
At w = 10.5, we have, l = 10.5 - w = 10.5 - 10.5 = 0
Therefore, the possible width is; w = 3.5
∴ The length, l = 10.5 - 3.5 = 7
The correct option is;
- B. Length = 7 cm, width = 3.5 cm
Learn more about finding the maximum volume of a cylinder here:
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