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In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possi

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In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possi  [#permalink]

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New post 27 May 2018, 07:54
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In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possible area of the rectangle ABCD?

A - 1000
B - 1200
C - 1250
D - 1500
E - 1600
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Re: In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possi  [#permalink]

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New post 27 May 2018, 09:11
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anamikarai wrote:
In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possible area of the rectangle ABCD?

A - 1000
B - 1200
C - 1250
D - 1500
E - 1600


In rectangle ABCD, lengths of AB=CD and AD=BC
Hence, AB+BC+CD=100 => 2AB+CD=100
AB cannot be equal to 50, since it will not be a rectangle then as 2AB=100 => CD=0. Not possible.
AB<50.
If AB=49, CD=100-2x49=2
Area of rectangle=49x2=98
Going on like this, if AB=25, CD=100-2x25=50
Area=25x50=1250
If AB=24, CD=100-2x24=52
Area=24x52=1248. Observe that the area is decreasing from 1250 obtained earlier.
Hence max. area can be 1250.
Answer C.
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Re: In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possi  [#permalink]

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New post 27 May 2018, 10:39
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Let AB=x and BC=y. So 2x+y = 100 and x < 65. Also area = x.y => x*(100-2x)

Maximize 100x - 2x^2 which gives x = 25 and y = 50. So area is 1250
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Re: In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possi  [#permalink]

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New post 27 May 2018, 10:42
anamikarai wrote:
In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possible area of the rectangle ABCD?

A - 1000
B - 1200
C - 1250
D - 1500
E - 1600


If BC is the length of the rectangle(let's call it x), AB and CD is the breadth of the rectangle(let's call it y)

Now, the question stem can be interpreted as x < 65 | x + 2y = 100. We have to calculate the value of xy.

Of the answer options, Option C is possible when x=50 and y=25 and is the maximum possible area of rectangle ABCD.
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Re: In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possi  [#permalink]

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New post 04 Jun 2018, 06:57
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Here what we understand from problem that 2X+Y=100 and we have to maximize the Area= x*y
Area= (100-2x)*x
Differentiate this equation with respect to x and dA/dx= 100-4x=0 that implies x= 25 and, from 1st equation y= 50
Maximum Area= 25*50=1250
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Re: In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possi  [#permalink]

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New post 01 Jul 2018, 09:54
AB+BC+CD=100 => 2AB+CD=100

A cannot be 50 then sum would be greater than 100

if AB=25, CD=100-2x25=50
Area=25x50=1250
If AB=24, CD=100-2x24=52
Area=1248

So area for further values keep on decreasing

So 1250 is answer

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Re: In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possi  [#permalink]

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New post 01 Jul 2018, 12:37
anamikarai wrote:
In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possible area of the rectangle ABCD?

A - 1000
B - 1200
C - 1250
D - 1500
E - 1600



Let AB = CD = x & BC = y, we have 2x + y = 100

Area of Rectangle as A = x*y

Now 2x + y = 100, we need to maximize x & y to maximize A.

Hence Max A can be when 2x = y = 50

x = 25 & y = 50, we get A = 1250

Now we can do a quick check for value of A at x = 26, we get A = 1248 < 1250

Similarly check at x = 24, we get A = 1248 < 1250

Hence we can safely say that Max A = 1250


Answer C.


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Re: In a rectangle ABCD, AB<65. if AB+BC+CD=100, what is the maximum possi &nbs [#permalink] 01 Jul 2018, 12:37
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