The radius of a tank can not exceed 5m, and the height can not exceed 12 m. Is it possible to construct a tank with a volume of 900m^3? If it is possible, determine the minimum amount of material required to make the tank.

When i solve it, the radius exceeds the restrictions.. so i believe that it is not possible. Can someone verify this?

Let the height be h, and the radius r

V = pir^2h
900 = pir^2h
h = 900/(pir^2)

For the surface area I am including 2 circles,
SA = 2pir^2 + 2pi(r)(h)
= 2pir^2 + 2pir(900/pir^2)
= 2pir^2 + 1800/r

d(SA)/dr = 4pir - 1800/r^2 = 0 for a min of SA
4pir = 1800/r^2
r^3 = (1800/(4pi)
r = 5.2

yes, you are right, so proceed with the largest value of r allowed, r = 5
then h = 900/(25pi) = 11.459 which is within the given restriction.

Plug in those values in my SA equation.

Well, constructing a tank with a volume of 900m^3 and avoiding the restrictions is quite a tall order. Or, well, maybe not so tall in this case. Anyway, let's see if we can figure this out.

The volume of a cylinder, which is the shape of a tank, is given by the formula V = πr^2h, where V is the volume, r is the radius, and h is the height.

So, we have the equation 900 = πr^2h.

Taking the restrictions into consideration, we have r ≤ 5m and h ≤ 12m.

Now, let's assume the radius is 5m and see what happens. Plugging it into the equation, we get:

900 = π(5)^2h
900 = 25πh
h = 900/(25π)
h ≈ 14.33m

Oh no, the height exceeds the limit of 12m! Looks like we can't construct a tank with a volume of 900m^3 given these restrictions.

So, to answer your question, it is not possible to construct a tank with a volume of 900m^3 while satisfying the restrictions. As for the minimum amount of material required, well, it seems we don't even need to worry about that in this case. Sorry to burst your tank-building bubble!

To determine if it is possible to construct a tank with a volume of 900m^3 within the given restrictions, we can use the formula for the volume of a cylinder: V = πr^2h.

Let's assume the radius of the tank is r and the height is h.

Since the radius cannot exceed 5m, we have r ≤ 5.

Similarly, since the height cannot exceed 12m, we have h ≤ 12.

We are given that the volume of the tank is 900m^3, so V = 900.

Plugging these values into the formula, we get:

900 = πr^2h.

Since the volume is fixed at 900m^3, we need to minimize the amount of material required to make the tank. This corresponds to minimizing the surface area of the tank.

The surface area of the tank can be given by the formula: A = 2πr^2 + 2πrh.

To minimize the surface area, we can differentiate the surface area formula with respect to r, set the derivative equal to zero, and solve for r.

Differentiating the surface area formula, we get:

dA/dr = 4πr + 2πh(dr/dr) = 4πr + 2πh.

Setting this equation equal to zero, we have:

4πr + 2πh = 0.

Solving for r, we get:

r = -h/2.

Since r represents the radius of the tank, it cannot be negative. Therefore, we can conclude that it is not possible to construct a tank with a volume of 900m^3 within the given restrictions.

Thus, your solution is correct. It is not possible to construct a tank with a volume of 900m^3 while also satisfying the given restrictions.

To determine if it is possible to construct a tank with a volume of 900m^3 while adhering to the given restrictions, we can use the formula for the volume of a cylinder: V = πr^2h, where V is the volume, r is the radius, and h is the height.

Let's start by assuming that it is possible. We can set up the equation and solve for the unknown values:

V = 900m^3
r ≤ 5m
h ≤ 12m

Substituting the given volume into the equation:

900 = πr^2h

Now, let's solve for the minimum amount of material required to make the tank. The tank's surface area is important in this case, as it represents the amount of material needed.

Surface Area = 2πrh + 2πr^2

To find the minimum surface area, we can minimize this expression.

First, express the surface area in terms of a single variable. Since the height (h) is already bounded (h ≤ 12m), we can express the surface area as a function of the radius (A(r)).

A(r) = 2πrh + 2πr^2

We can determine the minimum value of A(r) by taking the derivative with respect to r and setting it equal to zero:

A'(r) = 0

A'(r) = 2πh + 4πr = 0

Solving for r:

4πr = -2πh

r = -h/2

Since the radius (r) must be non-negative, we discard this solution.

Therefore, it is not possible to construct a tank with a volume of 900m^3 while adhering to the restrictions of a radius not exceeding 5m and a height not exceeding 12m.