Math Problem Statement

Assume that the height of your cylinder is 4 inches. Consider A as a function of r, so we can write that as A(r) = 2πr^2 + 8πr. What is the domain of A(r)? In other words, for which values of r is A(r) defined?

Solution

Let's break down the problem step by step:

Part a:

The surface area A(r)A(r) of a cylinder is given by the equation:

A(r)=2πr2+8πrA(r) = 2 \pi r^2 + 8 \pi r

Here, rr is the radius of the cylinder, and we are tasked with finding the domain of A(r)A(r).

Step 1: Determine the domain

  • The domain of a function is the set of all possible input values (here, rr) for which the function is defined.
  • Since the equation contains r2r^2 and a linear term rr, we need to ensure that rr is non-negative because the radius of a cylinder cannot be negative.
  • Therefore, the domain of A(r)A(r) is all r0r \geq 0.

Answer for Part a: The domain of A(r)A(r) is r0r \geq 0.


Part b:

We are asked to solve for rr as a function of AA, meaning we need to find the inverse function, r(A)r(A).

The equation we need to solve is:

A=2πr2+8πrA = 2 \pi r^2 + 8 \pi r

Step 2: Simplifying the equation

Divide both sides by 2π2\pi to make the quadratic term easier to manage:

A2π=r2+4r\frac{A}{2\pi} = r^2 + 4r

Now rearrange this as a quadratic equation in standard form:

r2+4rA2π=0r^2 + 4r - \frac{A}{2\pi} = 0

Step 3: Solve using the quadratic formula

The quadratic formula is:

r=b±b24ac2ar = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In our case, a=1a = 1, b=4b = 4, and c=A2πc = -\frac{A}{2\pi}.

Substitute these into the formula:

r=4±424(1)(A2π)2(1)r = \frac{-4 \pm \sqrt{4^2 - 4(1)\left(-\frac{A}{2\pi}\right)}}{2(1)}

Simplify the discriminant:

r=4±16+2A/π2r = \frac{-4 \pm \sqrt{16 + 2A/\pi}}{2}

Now divide the terms:

r=4±16+2A/π2r = \frac{-4 \pm \sqrt{16 + 2A/\pi}}{2}

This gives us two potential solutions. Since the radius must be non-negative, we take the positive root:

r=4+16+2A/π2r = \frac{-4 + \sqrt{16 + 2A/\pi}}{2}

Thus, the inverse function is:

r(A)=4+16+2A/π2r(A) = \frac{-4 + \sqrt{16 + 2A/\pi}}{2}


Part c:

Now, we're asked to evaluate rr when A=150A = 150 square inches.

Substitute A=150A = 150 into the equation for r(A)r(A):

r(150)=4+16+2(150)/π2r(150) = \frac{-4 + \sqrt{16 + 2(150)/\pi}}{2}

I'll calculate the exact value for r(150)r(150).For part c, when the surface area A=150A = 150 square inches, the radius rr is approximately:

r3.28 inchesr \approx 3.28 \text{ inches}

Let me know if you'd like further details or have any questions.

Here are 5 related questions to expand your understanding:

  1. How would the surface area formula change if the height of the cylinder were different?
  2. Can the quadratic formula be used for all values of AA, or are there any limitations?
  3. How does the domain of the radius function r(A)r(A) change as the surface area increases?
  4. What is the interpretation of the negative root in the quadratic formula in this context?
  5. How would the result change if the surface area was doubled?

Tip: Always check if simplifying the problem (e.g., dividing by common factors like 2π2\pi) helps make equations easier to solve!

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Math Problem Analysis

Mathematical Concepts

Algebra
Surface Area of Cylinders
Quadratic Equations

Formulas

Surface area of a cylinder A(r) = 2πr^2 + 8πr
Quadratic equation formula: r = [-b ± √(b² - 4ac)] / 2a

Theorems

Quadratic Formula

Suitable Grade Level

Grades 9-12