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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-example-1
User Evalarezo
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1 Answer

5 votes

Answer:

x = 140 y = 105 x · y = 14700 m²

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

User Rajh
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5.0k points