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

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

Mathematics
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the optimal measurements?
dimensions of rectangle are 5 and 12
how did you get that? :(

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iwas just about to say 5 and 12.... lol
you're both gods to me haha
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
so why is there a quadratic equation in the problem?
clutter probably :) maybe its part of the next problem
\[2(a+b)=34\quad,\quad\sqrt{a^2+b^2}=13\] solve this system of equations
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
thanks to both of you, but does the quadratic equation have anything to do with the problem
it can.... if you dont know the limitations of the other stuff, that can help you to find a length...maybe
thank you again, and nikvist too
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.
The quadratic equation is there because you need to find the value of the width using the quadratic equation.
\[x ^{2}+(17-x)^{2}=13^{2}=169\]
\[x ^{2}+289-34x+x ^{2}=169\] \[2x ^{2}-34x+289=169\] \[2x ^{2}-34x+120=0\] \[x ^{2}-17x+60=0\] Did you follow
You can factor solve it getting (x-5)(x-12)=0 x=5 and 12

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