anonymous
  • anonymous
perimeter of rectangle = 34 length of diagonal = 13 width = x derive x^2 - 17x + 60 = 0 find the dimensions of the rectangle
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
the optimal measurements?
nikvist
  • nikvist
dimensions of rectangle are 5 and 12
anonymous
  • anonymous
how did you get that? :(

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More answers

amistre64
  • amistre64
iwas just about to say 5 and 12.... lol
anonymous
  • anonymous
you're both gods to me haha
amistre64
  • 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
anonymous
  • anonymous
so why is there a quadratic equation in the problem?
amistre64
  • amistre64
clutter probably :) maybe its part of the next problem
nikvist
  • nikvist
\[2(a+b)=34\quad,\quad\sqrt{a^2+b^2}=13\] solve this system of equations
amistre64
  • 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
anonymous
  • anonymous
thanks to both of you, but does the quadratic equation have anything to do with the problem
amistre64
  • amistre64
it can.... if you dont know the limitations of the other stuff, that can help you to find a length...maybe
anonymous
  • anonymous
thank you again, and nikvist too
radar
  • 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.
anonymous
  • anonymous
The quadratic equation is there because you need to find the value of the width using the quadratic equation.
radar
  • radar
\[x ^{2}+(17-x)^{2}=13^{2}=169\]
radar
  • 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
radar
  • radar
You can factor solve it getting (x-5)(x-12)=0 x=5 and 12

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