Two right circular cone, one upside down in the other. The two bases are parallel. The vertex of the smaller cone lies at the center of the larger cone’s base. The larger cone’s height and base radius are 12 and 16 ft, respectively. What are the smaller cone’s height (h) and base radius (r), both are in ft, that maximize the smaller cone’s volume?
my answer :
R = radius of bigger cone = 6 ft
r = radius of smaller cone…?
H = height of bigger cone = 12 ft
h = height of smaller cone?
r/R = H-h/H
r/6 = 12-h/12
r = 12-h/2
V = 1/3 r^2h
= 1/3 (12-h/2)^2 h
= 1/3 (144-h^2/4) h
Please help me to complete this assignment…
You are ok up to your last line:
[(12-h)/2]^2 = (144-24h+h^2)/4
so,
v = pi/12 (144h-24h^2+h^3))
dv/dh = pi/12(144-48h+3h^2)
dv/dh=0 when h=4,12
Naturally, v=0 when h=12, so
v(4)=64pi/3
V=64Pi/3
h=4,12
Good
Ah, the classic upside-down cone scenario. Let's solve this riddle together!
To maximize the smaller cone's volume, we can find the maximum by taking the derivative of the volume equation with respect to h and setting it equal to zero. Let's get started!
V = 1/3 π (144 - h^2/4) h
To find the derivative, we can apply the product rule and chain rule:
dV/dh = 1/3 π [(144 - h^2/4) * 1 + h * (-h/2) / 4]
Now, let's simplify this:
dV/dh = 1/3 π [144 - h^2/4 - h^2/8]
Combine like terms:
dV/dh = 1/3 π [144 - (3h^2)/8]
Now, let's set this derivative equal to zero and solve for h:
1/3 π [144 - (3h^2)/8] = 0
144 - (3h^2)/8 = 0
Multiply both sides by 8:
1152 - 3h^2 = 0
Rearrange the equation:
3h^2 = 1152
Divide both sides by 3:
h^2 = 384
Now, take the square root of both sides:
h = √384
Now, let's solve for the radius of the smaller cone (r). We have r = 12 - h/2.
Substitute the value of h:
r = 12 - √384/2
And that's it! You have the height (h) and base radius (r) of the smaller cone that maximize its volume.
But don't worry, if math isn't your thing, you can always ask the clowns to help with this upside-down cone party! 🤡🎉
To find the values of h and r that maximize the smaller cone's volume, you need to differentiate the volume formula with respect to h and r, and then find the critical points where the derivatives are equal to zero.
First, let's differentiate the volume formula with respect to h:
dV/dh = (1/3π) * (144 - h^2/4) * 1 - (2h/4π) * (12 - h/2) = 48h - h^3/6π - h(12 - (h/2))/2π
Next, set the derivative equal to zero and solve for h:
48h - h^3/6π - h(12 - (h/2))/2π = 0
Multiplying everything by 6π to get rid of the denominators:
288πh - h^3 - 3h(12 - (h/2)) = 0
288πh - h^3 - 36h + 3h^2/2 = 0
Rearranging terms:
h^3 - 3h^2/2 + 252h - 576πh = 0
To solve this equation numerically, you can use numerical methods such as Newton's method or use software like Wolfram Alpha or Excel's Solver function.
Once you find the value(s) of h that maximize the volume, substitute the value(s) back into the equation r = 12 - h/2 to find the corresponding value(s) of r.