# 1. A man estimates that he can paint his house in x hours of his spare time if he buys equipment...

## Question:

1. A man estimates that he can paint his house in x hours of his spare time if he buys equipment costing {eq}100+\frac{5,000}{x^{2}} {/eq} dollars, and that his spare time is worth $2 per hour.

How many hours should he take (in order to minimize his total costs)?

2. Use Newton's method to estimate {eq}\sqrt{10} {/eq} to the nearest one one-hundredth. Show your method and work.

## Cost Minimization and Linearization:

In this problem, we will minimize a cost function. As with any other function, we can find the minimum value of a function by setting the derivative equal to zero, then solving the resulting function.

We will also use a linearization technique to approximate the value of a function. If we know the exact value of the function {eq}f(x) {/eq} at the point {eq}x = a, {/eq} and the derivative {eq}f'(a) {/eq} exists as well, then the linearization of the function is given by the formula {eq}y = f(a) + f'(a) (x - a). {/eq}

## Answer and Explanation:

1. Working for {eq}x {/eq} hours, his spare time would come to a cost of {eq}2x {/eq} dollars. Adding this to the cost of the equipment gives a total cost function

{eq}C(x) = 2x + 100 + \displaystyle\frac{5000}{x^2} = 2x + 100 + 5000 x^{-2}. {/eq}

The derivative of this function is

{eq}C'(x) = 2 - 10000x^{-3}. {/eq}

Setting this equal to zero gives

{eq}\begin{eqnarray*}2 - 10000x^{-3} & = & 0 \\ \\ 10000x^{-3} & = & 2 \\ \\ x^3 & = & 5000 \\ \\ x & = & \sqrt[3]{5000} \approx 17.1 \: \mathrm{hours} \end{eqnarray*} {/eq}

Therefore {eq}17.1 {/eq} hours of work will minimize the total cost of the job.

2. Newton's method is used to approximate the zero of a function. We wish to estimate {eq}\sqrt{10}. {/eq} We will linearize the function {eq}f(x) = \sqrt{x} {/eq} at the point {eq}x = 9. {/eq} The derivative of {eq}f {/eq} is {eq}f'(x) = \displaystyle\frac12 x^{-1/2}, {/eq} so {eq}f'(9) = \displaystyle\frac12 9^{-1/2} = \frac16. {/eq}The linearization of f at {eq}x = 9 {/eq} is

{eq}y = f(9) + f'(9) (x - 9) = 3 + \displaystyle\frac16 (x - 9) = \frac{x}{6} + \frac32. {/eq}

Setting {eq}x = 10 {/eq} gives the estimate {eq}\sqrt{10} = \displaystyle\frac{10}{6} + \frac32 \approx 3.17. {/eq} A calculator gives the approximation {eq}\sqrt{10} \approx 3.16 {/eq} to two decimal places.

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from Math 104: Calculus

Chapter 9 / Lesson 3