a. A closed cylindrical can is to hold 1000cm^3 of liquid. How should we choose the height and radius of this can to minimize the amount of material need to manufacture the can?
b.a manufacturer wants to design a open box having a square base and surface of 108cm^2. What dimensions will produce a box with minimum volume?
πr^2 h = 1000, so h = 1000/(πr^2)
surface area is
a = 2πr^2 + 2πrh
= 2πr^2 + 2000/r
da/dr = 4πr - 2000/r^2
= 4(πr^3 - 500)/r^2
da/dr=0 when r = ∛(500/π)
thus, h = 100/∛(250π)
Do the other along the same way.
dinesh bro...
the b is the same as the a
h=234.43
-1+x^2
a. To minimize the amount of material needed to manufacture the can, we need to find the dimensions (height and radius) that will result in the smallest surface area. Since the can is closed and cylindrical, its surface area is given by:
S = 2πrh + πr^2,
where S is the surface area, r is the radius, and h is the height of the can.
To minimize S, we need to minimize the equation mentioned above. However, there is a constraint given that the can should hold 1000 cm^3 of liquid. This means the volume of the can should be 1000 cm^3:
V = πr^2h = 1000.
We need to express h in terms of r from the volume equation, so we have:
h = 1000 / (πr^2).
Substituting this expression for h in the surface area equation, we get:
S = 2πr(1000 / (πr^2)) + πr^2 = 2000 / r + πr^2.
To minimize S, we can take the derivative of S with respect to r, set it equal to zero, and solve to find the critical points. Taking the derivative gives:
dS/dr = -2000 / r^2 + 2πr = 0.
Simplifying this equation, we have:
2πr = 2000 / r^2.
Multiplying both sides by r^2:
2πr^3 = 2000.
Solving for r, we find:
r^3 = 1000 / π.
Taking the cube root of both sides, we get:
r = (1000 / π)^(1/3).
Once we have the radius, we can substitute this value back into the volume equation to find the corresponding height:
h = 1000 / (πr^2).
Therefore, to minimize the amount of material needed to manufacture the can, we should choose the height as 1000 / (πr^2) and the radius as (1000 / π)^(1/3).
b. To find the dimensions that will produce a box with minimum volume, we need to consider the constraints given. The box has a square base, so its length and width are equal. Let's call the length/width of the square base "x" and the height of the box "h".
The volume of the box is given by:
V = x^2h.
We want to minimize the volume while also considering the constraint that the surface area is 108 cm^2. The surface area of the box consists of the area of the four sides and the area of the base. Since the box has a square base, the surface area equation becomes:
S = 4xh + x^2.
Now, we need to express h in terms of x from the surface area equation to eliminate one variable. Solving the surface area equation for h, we have:
h = (S - x^2) / 4x.
Substituting this expression for h in the volume equation, we get:
V = x^2[(S - x^2) / 4x] = (Sx - x^3) / 4.
To find the minimum volume, we can take the derivative of V with respect to x, set it equal to zero, and solve for the critical point. Taking the derivative gives:
dV/dx = (S - 3x^2) / 4 = 0.
Simplifying this equation, we have:
S = 3x^2.
Therefore, x^2 = S / 3.
Taking the square root of both sides, we get:
x = √(S / 3).
Once we have the value of x, we can substitute it back into the expression for h:
h = (S - x^2) / 4x.
Therefore, to produce a box with minimum volume, we should choose the length/width of the square base as √(S / 3) and the height as (S - x^2) / 4x.
answer
steve bro...
hw bout the b?
plz tell me...