an open topped cylinder has a volume of 125 cubic inches. determine the radius of the pot that will minimize it's surface area.
What I have so far...
radius =r
keight =h
V=πr²h
125/πr²=h
SA = πr² + 2πr(h)
= πr² + 2πr(125/πr²)
SA =πr² +250/r
so far so good:
A = pi r^2 + 250/r
dA/dr = 2pi r - 250/r^2
so, when is dA/dr = 0?
235.62
To determine the radius that will minimize the surface area of the open-topped cylinder with a volume of 125 cubic inches, we need to find the derivative of the surface area equation with respect to the radius, set it equal to zero, and solve for the radius.
Let's start by differentiating the surface area equation:
d(SA)/dr = d(πr²) + d(250/r)
= 2πr - 250/r²
Now, let's set the derivative equal to zero and solve for r:
2πr - 250/r² = 0
Multiplying through by r²:
2πr³ - 250 = 0
Rearranging the equation:
2πr³ = 250
Dividing by 2π:
r³ = 250/(2π)
Taking the cube root of both sides:
r = (250/(2π))^(1/3)
Simplifying:
r = (125/π)^(1/3)
Thus, the radius that will minimize the surface area of the open-topped cylinder is (125/π)^(1/3).
To find the radius of the pot that minimizes its surface area, we need to first express the surface area in terms of only the radius.
Given that the volume of the open-top cylinder is 125 cubic inches, we can write the equation for the volume as:
V = πr²h
Since the height of the cylinder is not provided, we need to eliminate it from the equation. We can solve for h in terms of r:
h = (125 / πr²)
Now we can substitute this expression for h into the equation for surface area:
SA = πr² + 2πr(h)
SA = πr² + 2πr(125 / πr²)
Simplifying further, we have:
SA = πr² + 250 / r
To minimize the surface area, we need to find the critical points of this equation. We can do this by taking the derivative with respect to r and setting it equal to zero:
dSA/dr = 2πr - 250 / r² = 0
To simplify this equation, we can multiply both sides by r²:
2πr³ - 250 = 0
Now isolate r:
2πr³ = 250
r³ = 250 / (2π)
r³ = 125 / π
Taking the cube root of both sides, we find:
r = (125 / π)^(1/3)
So, the radius of the pot that minimizes its surface area is equal to (125 / π)^(1/3).