## A community for students. Sign up today

Here's the question you clicked on:

## anonymous 5 years ago perimeter of rectangle = 34 length of diagonal = 13 width = x derive x^2 - 17x + 60 = 0 find the dimensions of the rectangle

• This Question is Closed
1. amistre64

the optimal measurements?

2. nikvist

dimensions of rectangle are 5 and 12

3. anonymous

how did you get that? :(

4. amistre64

iwas just about to say 5 and 12.... lol

5. anonymous

you're both gods to me haha

6. amistre64

the 13 diagonal means that there is only a few setups of legs to match.... it just so happens that a 5-12-13 triangle works

7. anonymous

so why is there a quadratic equation in the problem?

8. amistre64

clutter probably :) maybe its part of the next problem

9. nikvist

$2(a+b)=34\quad,\quad\sqrt{a^2+b^2}=13$ solve this system of equations

10. amistre64

you tend to use the derivative stuff to find the max or min area of the box..... but the 13 and 34 tend to limit you already

11. anonymous

thanks to both of you, but does the quadratic equation have anything to do with the problem

12. amistre64

it can.... if you dont know the limitations of the other stuff, that can help you to find a length...maybe

13. anonymous

thank you again, and nikvist too

14. radar

Here is how the $x ^{2}-17x+60$ Came into play. Let x = dimension L The perimeter is 34. 34-2x = 2y which is the other dimension doubled 2y. Divide that by 2 getting: 17-x= width. Now use that to get the diagonal.

15. anonymous

The quadratic equation is there because you need to find the value of the width using the quadratic equation.

16. radar

$x ^{2}+(17-x)^{2}=13^{2}=169$

17. radar

$x ^{2}+289-34x+x ^{2}=169$ $2x ^{2}-34x+289=169$ $2x ^{2}-34x+120=0$ $x ^{2}-17x+60=0$ Did you follow

18. radar

You can factor solve it getting (x-5)(x-12)=0 x=5 and 12

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy