Question 592700

Let L be the length and, W the width 
  We know that the perimeter of a rectangle is  2L + 2W , and that its area is  LW

 So, we have :  2L + 2W =  34  -->   L + W = 17
                LW = 66              LW=66

 At this point, there are many ways of solving it.
  
  First, let's use the easiest way but sometimes the longest . 
   find all the two numbers whose product is 66,   
we have: 1 and 66,  2 and 33,  3 and 22,   6 and 11
   Now, look for the two combination in which the two numbers add up to 17
 The only combination is 6 and 11 . So, the dimension are 6 and 11

 

The other way i am going to use is more complex.  
first you need to pick one equation,  write one variant in term of the other and plug in it in the other equation.  
 i chose the first equation, L+W=17  -->  L = 17 - W
By plugging it the other equation, we get  W( 17-W) = 66; Then , we solve for W
    W(17-W) = 66   -->  W^2 - 17W + 66 =0

  Now we use the quadratic formula.
W=[17 + sqrt(-17^2 -4( 1)(66))]/ 2   and    W=[17 - sqrt(-17^2 -4( 1)(66))]/
 =  (17 +  5)/2                              =(17 -  5)/2
 = 11                                        = 6
so,  the dimensions are 11 and 6.