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Re: PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100 [#permalink]
I think the area will be very much less than what's written in Q. B rather the maximum area possible should be 98.
given PQ + QR + RS = 100
As this is a rectangle PQ and RS must be same so we get
2(width) + QR = 100
now width should be less than 50 in order for QR to be positive
Now to have maximum area QR =2 PQ/RS= 49
Hence maximum area= 98

Where did I go wrong?
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Re: PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100 [#permalink]
1
PQRS is a rectangle. therefore PQ = RS = x, QR = SP = y
lets take PQ as x. So, PQ = RS = x

PQ+QR+RS = 100
2x+y = 100
y = 100 - 2x
AREA of rectangle = length*breadth
area = x*y
1250= x(100-2x)
solving it we get x= 25
therefore y = 50
AREA = 25*50 = 1250

Hence option C is correct as maximum area was asked.
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Re: PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100 [#permalink]
sanmitra wrote:
PQRS is a rectangle. therefore PQ = RS = x, QR = SP = y
lets take PQ as x. So, PQ = RS = x

PQ+QR+RS = 100
2x+y = 100
y = 100 - 2x
AREA of rectangle = length*breadth
area = x*y
1250= x(100-2x)
solving it we get x= 25
therefore y = 50
AREA = 25*50 = 1250

Hence option C is correct as maximum area was asked.


You cannot equate the area with Col. B and solve
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PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100 [#permalink]
why can't we assume the rectangle to be a square, since max area occurs in the case of a square? in this case, 3x=100 and the area will be >1250
Please point to where the gap in my logic is. Thanks a ton in advance
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Re: PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100 [#permalink]
Expert Reply
Essentially, above is calculated the slope which is 50. From that we can have the value of x and y. and calculate the max possible area.

The fact that it could be a square I do not think imply that much for my understanding
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Re: PQRS is a rectangle with PQ < 65 and PQ + QR + RS = 100 [#permalink]
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