Answer:
The answer is "138.10"
Step-by-step explanation:
Let r also be cone base radius and h = the cone height.
The radius for the original circle was on either side of the cone plied together = 7.
The hypotenuse is 7, as well as a right triangle, is formed. So:
[tex]\to h^2+r^2 = 49\\\\ \to r^2 = 49-h^2[/tex]
Calculating the volume of the cone:
[tex]V = (\frac{1}{3}) \pi r^2 h \\\\[/tex]
[tex]= (\frac{1}{3}) \pi (49 -h^2) h\\\\= (\frac{1}{3}) \pi 49h -h^3\\\\[/tex]
[tex]V'(h) = (\frac{1}{3}) \pi 49 - 3h^2\\\\[/tex]
Setting the derivative to zero and solving for h^2:
[tex]49-3h^2 = 0 \\\\ h^2 = \frac{49}{3} \\\\ r^2 = 49-\frac{49}{3} \\[/tex]
[tex]= \frac{ 147 -49}{3} \\\\= \frac{98}{3} \\\\[/tex]
Calculating the maximum volume:
[tex]= \frac{1}{3} \pi \frac{98}{3} \frac{7\sqrt{3}}{3} \\\\= \frac{1}{3} \times 3.14 \times 32.66 \times 4.04 \\\\= \frac{1}{3} \times 3.14 \times 32.66 \times 4.04 \\\\= 138.10[/tex]
