A open cylindrical wastepaper bin, of radius r cm and volume V cm^3, is to have a surface area of 5000cm^3.
a.Show that V=1/2 *r(5000-pir^2)
b.Calculate the maximum possible capacity of the bin.
let the height be h cm
SA = circular base + rectangular sleeve
= πr^2 + 2πrh
= 5000
2πrh = 5000 - πr^2
h = (5000-πr^2)/(2πr)
V= πr^2 h
= πr^2((5000-πr^2)/(2πr))
= (1/2)r(5000-πr^2) as asked for
= 2500r - πr^3/2
dV(dr) = 2500 - (3/2)π r^2 = 0 for a max of V
(3/2)πr^2 = 2500
r^2 = 5000/(3π)
r = √( 5000/(3π) ) = ....
Now go to your calculator, so far there was no need for it
check my algebra, I did not write it out on paper first
a. To show that V = 1/2 * r(5000 - πr^2), we can use the formula for the surface area of the open cylindrical wastepaper bin.
The formula for the surface area of a cylindrical bin is given by:
Surface Area = 2πrh + πr^2
Since the bin is open, we can ignore the top surface area (πr^2). Therefore, the surface area of the bin becomes:
Surface Area = 2πrh
Given that the surface area is 5000 cm^2, we have:
5000 = 2πrh
Now, we need to express the height h in terms of r and V.
The volume of a cylinder is given by the formula:
Volume = πr^2h
Therefore, h = V / (πr^2).
Substituting this value of h into the surface area equation, we get:
5000 = 2πr(V / (πr^2))
Simplifying further, we have:
5000 = 2V / r
Multiplying both sides by r, we get:
5000r = 2V
Finally, dividing both sides by 2, we obtain:
V = 5000r / 2
Simplifying further, we get:
V = 1/2 * 5000r
V = 1/2 * r(5000 - πr^2)
b. To calculate the maximum possible capacity of the bin, we need to find the maximum value of V. In order to do that, we can differentiate the equation in part (a) with respect to r and find the critical point where the derivative is zero.
Differentiating the equation V = 1/2 * r(5000 - πr^2) with respect to r, we get:
dV/dr = 1/2 * (5000 - 3πr^2)
Setting dV/dr equal to zero, we have:
0 = 5000 - 3πr^2
Rearranging the equation, we get:
3πr^2 = 5000
Dividing both sides by 3π, we obtain:
r^2 = 5000 / (3π)
Taking the square root of both sides, we get:
r = √(5000 / (3π))
Substituting this value of r into the equation for V that we found in part (a), we can calculate the maximum possible capacity of the bin:
V = 1/2 * r(5000 - πr^2)
V = 1/2 * (√(5000 / (3π))) * (5000 - π(√(5000 / (3π))))^2
By evaluating this expression, we can find the maximum possible capacity of the bin.
To derive the formula for the volume of the wastepaper bin, we need to relate it to the given surface area. Let's break it down step by step:
a. To find the formula for V, we begin with the surface area of the cylindrical bin:
Surface Area = 2πr² + 2πrh
Since the bin is open, we don't have a top surface. Hence, we only consider the curved surface area.
Now, we are given that the surface area is 5000 cm²:
5000 cm² = 2πr² + 2πrh
We need to eliminate 'h' from this equation. We can do so by relating it to V, the volume.
The volume of a cylinder is given by:
V = πr²h
Solving for h, we get:
h = V / (πr²)
Substituting this expression for 'h' into the equation for surface area:
5000 cm² = 2πr² + 2πr * (V / (πr²))
Simplify:
5000 cm² = 2πr² + (2V / r)
Rearranging:
2πr² = 5000 cm² - (2V / r)
πr² = 2500 cm² - (V / r)
Multiply through by 2:
2πr² = 5000 cm² - (2V / r)
Subtract πr²:
πr² = 5000 cm² - (2V / r)
Finally, divide by πr² to solve for V:
V = (πr² * 5000 cm²) / (πr²) - 2πr²
Simplify further:
V = 5000 cm² - 2πr²
V = 5000 - 2πr² (since πr² simplifies to 1)
V = 5000 - 2πr²
V = 5000 - 2πr^2 (final simplified form)
b. To calculate the maximum possible capacity (volume) of the bin, we need to determine the maximum value of V.
To do this, we differentiate the volume equation with respect to 'r' and find where the derivative is zero:
dV/dr = -4πr
Set dV/dr = 0 to find the critical point:
-4πr = 0
From here, we can see that r = 0 is a critical point. However, since the radius represents a physical length, it cannot be zero.
So, there are no critical points, and we conclude that the volume V is always decreasing as r increases.
Since we're looking for the maximum capacity, it means r must be as small as possible. Therefore, we choose the minimum allowed value for r, which is zero.
Substituting r = 0 into the volume equation:
V = 5000 - 2π(0)^2
V = 5000 cm^3
Thus, the maximum possible capacity of the bin is 5000 cm³.