A manufacturer needs to make a cylindrical can that will hold 1.5 liters of liquid. Determine the dimensions of the can that will minimize the amount of material used in its construction.
πr^2 h = 1500 cm^3
h = 1500/(πr^2)
The material used has area
A = 2πr^2 + 2πrh = 2πr^2 + 2πr*1500/(πr^2) = 2πr^2 + 3000/r
dA/dr = 4πr - 3000/r^2
so now just find where dA/dr = 0
Well, if we're trying to minimize the amount of material used in the construction of a cylindrical can, perhaps we should consider making an invisible can. That way, we won't be using any material at all! Problem solved! 😉 But on a serious note, let's find the dimensions of the can that will do the trick!
To minimize the amount of material used, we need to minimize the surface area of the can while still holding 1.5 liters of liquid. The surface area of a cylinder is given by the formula A = 2πrh + 2πr^2, where r is the radius and h is the height of the cylinder.
Since we want to hold 1.5 liters of liquid, we need to convert that to the volume of a cylinder using the formula V = πr^2h. Let's say the radius of the can is r and the height is h.
Given that V = 1.5 liters = 1500 cm^3, we can rearrange the volume formula to solve for h: h = V / (πr^2).
Now, substitute this value of h into the surface area formula: A = 2πr(V / (πr^2)) + 2πr^2.
Simplifying, we get A = 2(V / r) + 2πr^2.
To minimize A, we need to take the derivative of A with respect to r and set it equal to 0. Solving for r will give us the radius that minimizes the surface area.
All this math talk might make you dizzy, but don't worry, I'll be the clown supplying the fun while you handle the calculations! Let's crunch those numbers! 🎉🤡
To determine the dimensions of the can that will minimize the amount of material used, we can use the concept of optimization.
Let's start by defining some variables:
- Let r be the radius of the can.
- Let h be the height of the can.
- Let V be the volume of the can, which is given as 1.5 liters or 1500 cm³ (since 1 liter = 1000 cm³).
- Let A be the surface area of the can, which represents the amount of material used.
Since the can is cylindrical, we can use the formulas for the volume and surface area of a cylinder:
Volume of a cylinder: V = πr²h
Surface area of a cylinder: A = 2πrh + 2πr²
Now, we need to minimize the surface area A while keeping the volume V constant.
Step 1: Express the surface area in terms of one variable.
Using the volume equation, we can express h in terms of V and r:
V = πr²h
h = V / (πr²)
Substituting this expression for h into the surface area equation, we get:
A = 2πr(V / (πr²)) + 2πr²
Simplifying, we have:
A = (2V / r) + 2πr²
Step 2: Find the derivative of A with respect to r.
To find the critical points of A, we take the derivative of A with respect to r and set it equal to zero:
dA/dr = -2V / r² + 4πr
Step 3: Solve for r.
Setting dA/dr equal to zero, we get:
-2V / r² + 4πr = 0
-2V / r² = -4πr
r³ = V / 2π
r = (V / 2π)^(1/3)
Step 4: Find the corresponding height h.
We can substitute the value of r we found into the equation for h:
h = V / (πr²)
Step 5: Calculate the final dimensions of the can.
Now that we have both the radius r and the height h, we can calculate the dimensions of the can.
To determine the dimensions of the can that will minimize the amount of material used in its construction, we can start by setting up the mathematical model for the problem.
Let's assume the height of the cylindrical can is "h" and the radius of the circular base is "r." The volume of a cylinder is given by the formula V = πr²h, where π is approximately equal to 3.14159.
Given that the can needs to hold 1.5 liters of liquid, which is equal to 1500 cubic centimeters (since 1 liter = 1000 cubic centimeters), we have:
V = 1500
Substituting the formula for the volume in terms of r and h, we can rewrite the equation as:
πr²h = 1500
Now, we want to minimize the amount of material used, which is directly proportional to the surface area of the can. The surface area of the cylindrical can is the sum of the area of the circular base (A₁ = πr²) and the lateral surface area (A₂ = 2πrh).
So the total surface area (A) is given by:
A = A₁ + A₂
= πr² + 2πrh
To minimize the surface area, we can use the calculus method by finding the critical points. For that, we need to differentiate A with respect to r and h, by considering the volume equation πr²h = 1500 as a constraint.
Differentiating A with respect to r, we get:
dA/dr = d(πr²)/dr + d(2πrh)/dr
= 2πr + 2πh(dr/dr)
= 2πr + 2πh(1)
= 2πr + 2πh
Since dh/dr is dr/dr = 1
Similarly, differentiating A with respect to h, we get:
dA/dh = d(πr²)/dh + d(2πrh)/dh
= 0 + 2πr(dh/dh)
= 2πr
Now, we have two equations:
dA/dr = 2πr + 2πh
dA/dh = 2πr
We need to find the values of r and h that satisfy these two equations and the volume constraint. So, let's solve this system of equations.
From the equation dA/dr = 2πr + 2πh, we can rearrange and substitute πr²h = 1500:
2πr + 2πh = 0 (since dA/dr = 0 at critical points)
2πr + 2π(1500/(πr²)) = 0 (substituting πr²h = 1500)
r + 1500/r² = 0
Multiplying both sides by r² to eliminate the fractional term, we have:
r³ + 1500 = 0
Now, with the equation dA/dh = 2πr, we can substitute the value of r from the previous equation:
2πr = 0
2π(-1500/r²) = 0
-3000/ r² = 0
Simplifying these equations, we find that r = -10 and h = 3/5. However, since r cannot be negative in this context, we discard the negative value of r.
Therefore, the dimensions of the can that will minimize the amount of material used in its construction are:
Radius (r) = 10 cm
Height (h) = 3/5 cm