I am just so stumped on how to solve either of these so if you could so your work and explain what you did to arrive at your answer it would be greatly appreciated! Thank you!!

There's not much to compute here so this is just some general reasoning. Imagine you have a cylinder, what is the least area this cylinder could have?

0, right?

If the area was lower there would be no cylinder. So which is the smallest radius for the Area to be at least 0? Again, it is 0. Now you should probably assume that with radius 0 you would still not have a cylinder, but for any radius bigger than 0 there will be a cylinder, no matter how small it would appear. Therefor we say that the domain is r > 0.

b) Write r as a function of A.

Using the hints we get, the advice is to consider this problem as a quadratic equation [tex]2\Pi r^{2} +16\Pi r - A = 0[/tex]. I would simplify this expression further as,

[tex]r^{2} +8 r - \frac{A}{2\Pi} = 0[/tex]

The roots formula of the quadratic equation is well known. In this equation we get two roots for r out of which one is necessarily negative and we will ignore it for the reason given in part a. The positive root expression will be,

[tex]r = \frac{-8}{2} + \sqrt{(\frac{-8}{2})^{2} + \frac{A}{2\Pi}} = -4 + \sqrt{16 + \frac{A}{2\Pi}}[/tex]

Even with Area (A) being zero, this root will be zero because the root of 16 is 4, so the input value should be A > 0 (strictly bigger than 0).

c) Evaluate r(250)

This is just a formality. Insert 250 into the root expression at the end of part b.

[tex]r(250) = -4 + \sqrt{16+\frac{250}{2\Pi} } \approx -4 + 7.47\approx 3.47[/tex]

The radius for a cylinder with height 8 inches and area of 250 inches² would therefor be 3.47 inches. Doesn't sound like much but has a lot to do with the height.

Ok, that's all. If you have any questions about this explanation, just leave a comment. :)

semsee45 semsee45 · Answer 2 · 2024-01-15T22:07:39+01:00

Answer:

[tex]\textsf{(a)}\quad \textsf{Domain:}\;\;(0, \infty)[/tex]

[tex]\textsf{(b)}\quad r(A)=-4 + \sqrt{16+\dfrac{A}{2\pi}}[/tex]

[tex]\textsf{(c)}\quad r=3.47\; \sf inches\;(2\;d.p.)[/tex]

Step-by-step explanation:

Part a

The formula for the surface area of a cylinder is:

[tex]\boxed{\begin{array}{l}\underline{\textsf{Surface Area of a Cylinder}}\\\\A=2\pi rh+2\pi r^2\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$A$ is the surface area.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius of the circular base.}\\\phantom{ww}\bullet\;\textsf{$h$ is the height.}\end{array}}[/tex]

If the height of the cylinder is 8 inches, then we can substitute h = 8 into the formula to create an equation for A in terms of r:

[tex]A = 2\pi r^2 + 2\pi r(8)[/tex]

[tex]A = 2\pi r^2 + 16\pi r[/tex]

Since the surface area of a cylinder must be positive, the radius of a cylinder also has to be positive. Therefore, the values of r for which A(r) is defined are all positive real numbers, making the domain of A(r) (0, ∞).

[tex]\hrulefill[/tex]

Part b

To express the radius (r) as a function of A, we need to solve the equation for r.

Begin by rearranging the equation into a quadratic in the form ar² + br + c = 0:

[tex]2\pi r^2 + 16\pi r-A=0[/tex]

Now, we can use the quadratic formula to solve for r.

[tex]\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}[/tex]

In this case, the variable is r, so:

[tex]x = r[/tex]
[tex]a = 2 \pi[/tex]
[tex]b = 16 \pi[/tex]
[tex]c = -A[/tex]

Substitute the values into the quadratic formula:

[tex]r=\dfrac{-(16\pi)\pm\sqrt{(16\pi)^2-4(2\pi)(-A)}}{2(2\pi)}[/tex]

[tex]r=\dfrac{-16\pi\pm\sqrt{256\pi^2+8\pi A}}{4\pi}[/tex]

[tex]r=\dfrac{-16\pi}{4\pi}\pm\dfrac{\sqrt{256\pi^2+8\pi A}}{4\pi}[/tex]

[tex]r=-4\pm\dfrac{\sqrt{256\pi^2+8\pi A}}{4\pi}[/tex]

Rewrite the denominator by squaring and square rooting it:

[tex]r=-4\pm\dfrac{\sqrt{256\pi^2+8\pi A}}{\sqrt{16\pi^2}}[/tex]

Therefore:

[tex]r=-4\pm\sqrt{\dfrac{256\pi^2+8\pi A}{16\pi^2}}[/tex]

[tex]r=-4\pm\sqrt{\dfrac{256\pi^2}{16\pi^2}+\dfrac{8\pi A}{16\pi^2}}[/tex]

[tex]r=-4\pm\sqrt{16+\dfrac{A}{2\pi}}[/tex]

As r > 0, we take the positive solution only:

[tex]r=-4 + \sqrt{16+\dfrac{A}{2\pi}}[/tex]

[tex]\hrulefill[/tex]

Part c

To find the radius if the surface area is 250 square inches, substitute A = 250 into the equation for r:

[tex]r(250)=-4 + \sqrt{16+\dfrac{250}{2\pi}}[/tex]

[tex]r(250)=-4 + \sqrt{16+\dfrac{125}{\pi}}[/tex]

[tex]r(250)=3.4691857503...[/tex]

[tex]r(250)=3.47\; \sf inches\;(2\;d.p.)[/tex]

Therefore, the radius is 3.47 inches (rounded to two decimal places).