Given dV/dt = 0 , r=20 , h=2 , dr/dt = 0.5 . Find dh/dt

(r+(h/2))^3-(r-(h/2))^3 = 3hr^2 + (h^3)/4

dV/dt = 0 ---> 3(r^2)(dh/dt)+6hr(dr/dt)+(3/4)(h^2)(dh/dt) = 0

substitute values --> dh/dt = -40/401 cm/s

dV/dt = 0

V = (4/3) * pi * ((r + h/2)^3 - (r - h/2)^3)

a^3 - b^3 = (a - b) * (a^2 + ab + b^2)

(r + h/2)^3 - (r - h/2)^3 =>

(r + h/2 - r + h/2) * ((r + h/2)^2 + (r + h/2) * (r - h/2) + (r - h/2)^2) =>

h * (r^2 + rh + h^2 + r^2 - (1/4) * h^2 + r^2 - rh + h^2) =>

h * (3r^2 + (7/4) * h^2) =>

3h * r^2 + (7/4) * h^3

V = (4/3) * pi * (3hr^2 + (7/4) * h^3)

V = (4/3) * pi * (1/4) * (12hr^2 + 7h^3)

V = (pi/3) * (12hr^2 + 7h^3)

dV/dt = (pi/3) * (12 * (h * 2r * dr/dt + r^2 * dh/dt) + 21h^2 * dh/dt)

0 = (pi/3) * (12 * (2hr * dr/dt + r^2 * dh/dt) + 21h^2 * dh/dt)

0 = 12 * (2hr * dr/dt + r^2 * dh/dt) + 21h^2 * dh/dt

0 = 4 * (2hr * dr/dt + r^2 * dh/dt) + 7h^2 * dh/dt

0 = 8hr * dr/dt + 4r^2 * dh/dt + 7h^2 * dh/dt

0 = 8hr * dr/dt + (4r^2 + 7h^2) * dh/dt

r = 20 , h = 2 , dr/dt = 0.5 , dh/dt = ?

0 = 8 * 2 * 20 * 0.5 + (4 * 20^2 + 7 * 2^2) * dh/dt

0 = 160 + 4 * (20^2 + 7) * dh/dt

0 = 40 + 407 * dh/dt

-40 = 407 * dh/dt

-40/407 = dh/dt

For (a), consider that the tip of the cone and the entire circular base of the cone coincide with the surface of the sphere. That means that a line drawn from the center of the sphere to a point on the circular base is a radius of the sphere, with length 8 cm. It is also the hypotenuse of a right triangle that has as legs the radius of the circular base and a line drawn from the center of the circular base to the center of the sphere. You know that the radius of the circular base is r, so you can use the Pythagorean theorem to find the length of the other leg, which is part of the cone's height. The height of the cone above the center of the sphere is another radius of the sphere, having length 8 cm. Add those two values to get the total height of the cone. This will be h in terms of r; use algebra to rewrite it as r in terms of h.

Note that for any given radius (and assuming the tip of the cone is at the top of the sphere), there are two heights that satisfy these conditions; a short one, with the base above the center of the sphere, and a tall one, with the base below the center of the sphere. I assume we are dealing with the tall cone in all cases.

For (b), use the fact that V = (1/3)*(pi*r^2)*h, and use the expression for r in terms of h to rewrite V in terms of h only.

For (c), instead use the original expression for h in terms of r to rewrite V in terms of r only. V has "stationary value" when its derivative with respect to r is equal to zero. So take the derivative of V, set it equal to zero, and solve for r.

You can do (d) by taking a test value of r or by taking the second derivative.

Hello

V = 536 = wLh -> h = 536/(wL)

C = 2*w*L*5 + 3(2*w*h + 2L*h)

the base must in any case be quadratic, because 1. a square has a maximum area at minimum perimeter, and 2. a cube holds a maximum volume at minimum area. So an "elongated cube" is the second best structure for maximum volume. And in addition, the four sides have the same prize.

This makes the equation to

C = 10L^2 + 3*(2L*536/L^2 + 2L*536/L^2)

C = 10L^2 + 6432/L

C' = 20L - 6432/L^2 = 0

20L = 6432/L^2

20L^3 = 6432

L^3 = 6432/20

L^3 = 321,6

L = 6,85 cm

w = 6,85 cm

h = 536/L^2 = 11,42 cm

I hope that this is right so, or did you forget any other relation between w,L,h, and Volume or Area?

Regards

Do you mean:

"The sides of the box cost 3 cent/cm^2 and the top and bottom cost 5 cent/cm^2"

and not

"The sides of the box cost 3 cent/cm^3 and the top and bottom cost 5 cent/cm^3"?