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April 25, 2015

April 25, 2015

Posted by **Anonymous** on Monday, February 13, 2012 at 5:00pm.

- pre cal -
**Damon**, Monday, February 13, 2012 at 5:32pmx^2 + y^2 = diagonal^2 = d^2

x y = 4000 so y = 4000/x

d = x+10 so d^2 = x^2 + 20 x + 100

x^2 + 16*10^6/x^2 = x^2 + 20 x + 100

16*10^6 = 20 x^3 + 100 x^2

5 x^3 + 25 x^2 - 4*10^6 = 0

I do not see an easy way to solve that. Use iteration, x and 5x^3+25 x^2

x = 10^2 --> 5,250,000 too big

x = 80 --> 2,720,000 too small

x = 90 --> 3,847,500 getting close

x = 92 --> 4,333,568 very close

x = 91 --> 3,974,880 between 91 and 92

x = 91.5-> 4,039,610 between 91 and 91.5

so 91 to nearest foot

91^2 + y^2 = (91+10))^2

8281 +y^2 = 10201

y = 43.8 or 44

so

91 by 44

- pre cal -
**Reiny**, Monday, February 13, 2012 at 5:42pmlet one side be x

then the diagonal is x+10

let the third side by y

x^2 + y^2 = (x+10)^2

x^2 + y^2 = x^2 + 20x + 100

y^2 = 20x + 100

y = √(20x+100)

so the area of the triangle is half the rectangle area

(1/2)xy =(1/2)(4000)

xy = 4000

x√(20x+100) = 4000

x^2(20x+100) = 16000000

20x^3 + 100x^2 - 16000000 = 0

x^3 + 5x^2 - 800000 = 0

I must admit that I used Wolfram to solve this

http://www.wolframalpha.com/input/?i=x%5E3+%2B+5x%5E2+-+800000+%3D+0

**and I got x = 91.1947**

then y = 43.8622

and the diagonal would be 101.1947

check:

area of rectangle = xy = (91.1947)(43.8622) = 4000.00017 (close enough)

I will let you check if Pythagoras also works out, it does.

I don't know what method you would have to solve that cubic, since it does not factor, and therefore does not have rational roots.

Do you know Newton's Method ?