Fencing with the length of 840 m surrounds two identical playing fields with a common side. The fields can be side-by-side and separated by a fence, as shown in the figure. Find the dimensions in the maximum area of each field.

Fencing with the length of 840 m surrounds two identical playing fields with a common side The fields can be sidebyside and separated by a fence as shown in the class=

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Answer:

x = 140   y = 105                x · y = 14700 m²

Step-by-step explanation:

good question

it looks like 3 sides of length x

and 4 sides of length y

area 1 = area 2

and the length of all the sides sum to 840

  3x + 4y = 840      find the maximum area = 2· x · y

This is a calculus problem that I can't remember how solve

 if x = 0   y = 210

           if   y = 0   x = 280       280 / 2 = 140   the mid-point  the logical answer

if x = 140   y = 105                x · y = 14700 m²         this appears to be the max

if x = 135   y = 108.75            x · y = 14681.25 m²      smaller area

if x = 145   y = 101.25             x · y = 14681.25 m²      smaller area

I need an eqaution with a quadratic and take the first derivative

 y = (840 - 3x) / 4      xy = max area     y = max area/x

(max area) / x = (840 - 3x) / 4      solve for x

    max area  = x(840 - 3x) / 4

    max area  = [840x/4 - 3x²/4] '        

    max area  = 210x  -  3x²/4            take the first derivative  

    max area  = 210  -  6x/4               maximum occurs when slope = 0

           0   =  210 - 3x/2    

        3x/2 = 210

           x  = 210·2/3

           x  = 420 / 3

          x  = 140   y = 105

3(140) + 4(105) =   420 + 420  = 840    checks

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