mcsuds
  • mcsuds
Find the largest possible volume of a cone that fits inside a sphere of radius one.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mcsuds
  • mcsuds
@ganeshie8 @Hero
anonymous
  • anonymous
you get the volume of the sphere which is 4.18879
anonymous
  • anonymous
then you just have to find a volume of a cone smaller than that

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amilapsn
  • amilapsn
https://sketch.io/render/sketch55e116f6c4926.png
amilapsn
  • amilapsn
Use the above diagram and find an equation to represent the volume of the cone in terms of r. Then you can get the answer :-)
madhu.mukherjee.946
  • madhu.mukherjee.946
Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3? Let R = radius sphere r = base radius cone R + h = height cone V = volume cone _______ V = (1/3)πr²(R + h) By the Pythagorean Theorem: r² = R² - h² Plug into the formula for volume. V = (1/3)π(R² - h²)(R + h) = (1/3)π(R³ + R²h - Rh² - h³) Take the derivative and set equal to zero to find the critical points. dV/dh = (1/3)π(R² - 2Rh - 3h²) = 0 R² - 2Rh - 3h² = 0 (R - 3h)(R + h) = 0 h = R/3, -R But h must be positive so: h = R/3 Calculate the second derivative to determine the nature of the critical points. d²V/dh² = (π/3)(-2R - 6h) < 0 So this is a relative maximum which we wanted. Solve for r². r² = R² - h² = R² - (R/3)² = R²(1 - 1/9) = (8/9)R² Calculate maximum volume. V = (π/3)[(8/9)R²](R + R/3) = (8/27)πR³(4/3) = 32πR³/81 For R = 3 maximum volume is: V = 32π(3³)/81 = 32π(27)/81 = 32π/3
madhu.mukherjee.946
  • madhu.mukherjee.946
this was the question that came in our examination and this is how i solved it ....and this is similar to your problem only difference is that you gotta put 1 in place 3 (radius):))))))

Looking for something else?

Not the answer you are looking for? Search for more explanations.