Let $f(x)=e^{-1/x}$ . Then $f'(x)=\frac1{x^2}e^{-1/x}>0$ for all $x\ge2$ , so $f$ is strictly increasing. As $x\to\infty$ , $e^{-1/x}\to e^0=1$ , so $f$ is bounded above by 1. This is to say,

$e^{-1/x}$

and the integral of $\frac1{x^2}$ converges over the same domain, so this integral must also converge by comparison.

We have, by setting $y=-\frac1x$ ,

$\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}$

8 0

3 years ago

Austin began a repiping job with 26 yards, 2 feet, 5 inches of pipe. When he finished the job, he had 13 yards, 2 feet, 7 inches

Masteriza [31]

Your answer to the question is c

5 0

3 years ago

A furniture company manufactures desks and chairs. Each desk uses four units of wood, and each chair uses three units of wood. A

Troyanec [42]

Step-by-step explanation:

Although I cannot find any model or solver, we can proceed to model the optimization problem from the information given.

the problem is to maximize profit.

let desk be x

and chairs be y

400x+250y=P (maximize)

4x+3y<2000 (constraints)

according to restrictions y=2x

let us substitute y=2x in the constraints we have

4x+3(2x)<2000

4x+6x<2000

10x<2000

x<200

so with restriction, if the desk is 200 then chairs should be at least 2 times the desk

y=2x

y=200*2

y=400

we now have to substitute x=200 and y=400 in the expression for profit maximization we have

400x+250y=P (maximize)

80000+100000=P

180000=P

P=$180,000

the profit is $180,000

5 0

3 years ago