Question

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

Answer 1

the optimal measurements?

Answer 2

dimensions of rectangle are 5 and 12

Answer 3

how did you get that? :(

Answer 4

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

Answer 5

you're both gods to me haha

Answer 6

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

Answer 7

so why is there a quadratic equation in the problem?

Answer 8

clutter probably :) maybe its part of the next problem

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

thank you again, and nikvist too

Answer 14

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.

Answer 15

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

Answer 16

\[x ^{2}+(17-x)^{2}=13^{2}=169\]

Answer 17

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

Answer 18

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