perimeter of the rectangulatr room=240ft

perimeter of a rectangle=2*(l+b)

so 240=2*(l+b)so

l+b=120

so if l=x

b=120-x

area=x*(120-x)

A=x(120-x)

for maximum A dA/dx=0

dA/dx=d/dx(120x-x^2)

0=120-2x

x=60ft

length=60ft

breadth=120-60=60ft answer

P = 240 = 2L + 2W

A = L*W

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By theorem (or from calculus), we know the maximum area is a square, so: L=W.

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P = 2L + 2W = 4L = 240

L = 60

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So a 60 by 60 room is 3600 sq ft in area.

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We can test the area by testing another room with dimensions close to 60 by 60 and P = 240.

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Let's try 59 by 61. P = 2(59)+2(61) = 118+122 = 240. OK.

A = 59 * 61 = 3599, which is less than 3600.

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Let's try 58 by 62. P = 2(58) + 2(62) = 116 + 124 = 240. OK

A = 58 * 63 = 3596, which is less than 3600 and less than 3599.

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Let's try 57 by 63. P = 240. OK

A = 57 * 63 = 3591, which is less than 3600 and less than 3599 and less than 3596. The area is decreasing as the values get further apart.

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How about 1 by 119? P = 2(1) + 2(119) = 2 + 238 = 240.

A = 1*119 = 119, which is way less...

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Another way to look at this problem is the define the perimeter = 2C, where C = L+W.

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That way L = C-W, so the area will be (C-W) * W = CW - W^2.

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The maximum value of CW - W^2 is C^2/4, which occurs at W = C/2.

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So, if W = C/2, and C=L+W, then L=C/2, so we have a square.