Question

The sum of r and h for this cylinder is 12 cm.

a. Write down an equation for the area, A, of the curved surface in terms of r.
b. Find da/dr
c. Find the value ofr when the area of the curved surface is maximized.

Answer 1

a)

`A=2\pi(12r-r^(2))`

b)

`(dA)/(dr)=2\pi(12-2r)`

c) The value of r that maximizes A is r = 6cm

Step-by-step explanation:

We know that

`h+r=12`

We also want to find an expression for the curved area. This area is a rectangle with the length of the base equal to the circumference of the circle, and the height h.

The circumferece of a circle of radius r is:

`C=2\pi r`

The the area of the rectangle is base times height:

`A=C\cdot h=2\pi rh`

Now, since we want the area in terms of the radius, we can solve for h on relation we know between them:

`h+r=12\text{ Thus }h=12-r`

And now we can replace h in terms of x:

`A=2\pi r(12-r)`

And we can apply the distributive property to make de differentiation easier:

`\begin{gathered} A=2\pi(12r-r^2) \\ \end{gathered}`

And that's the answer to item a.

For item b, we need to differentiate. We can do this first using this property:

`(d)/(dx)(a\cdot f(x))=a(d)/(dx)f(x)`

We can take out the 2π outside the differentiation.

Next, we can use the power rule:

`2\pi\cdot(d)/(dr)(12r-r^2)=2\pi(12-2r)`

Thus, this is the answer to b

Finally, the item c ask us to find the radius when the area is maximized. When we have a derivative of a function, the x's where the derivative is zero, in the function there is a minimum or ma maximum.

In this case, we have a quadratic equation, and we know that this functions have only one maximum or only one minimum. SInce the principal coeficcient is negative (the one multiplying the r²) the parabolla ha it's "arms down" and in the vertex there is a maximum.

Then, let's find the sero of the derivative:

`\begin{gathered} 0=2\pi(12-2r) \\ 0=12-2r \\ 2r=12 \\ r=6 \end{gathered}`

The radius for which the curved area is maximum is 6cm. Note that this means that

`\begin{gathered} \begin{cases}h=12-r{} \\ r=6{}\end{cases} \\ Then, \\ h=12-6=6 \end{gathered}`

And the maximum is when r = h = 6. This makes the curved surface a square, and the square in the quadrilateral that maximizes the area for a given perimeter.