A rectangular playground with an area of 2400 square meters is bordere

Refer attached image for nomenclature

In the image playground is in green color, concrete wall is in black color and walking path is in orange color
Length of the walking path = a + b + a = 2a + b = 140 => b = 140 -2a

Area of Park = a*b = 2400
Using b=140-2a we get
a*(140-2a) = 2400
=> a*2*(70-a) = 2400
Diving by 2 get
a * (70-a) = 1200

Now there are two ways of solving this
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Method 1: Rearranging Terms

a * (70-a) = 1200 = 30 * 40 = 30 * (70 - 30) => a = 30
or, a * (70-a) = 1200 = 40 * 30 = 40 * (70 - 40) => a = 40

Method 2: Quadratic Equation

a * (70-a) = 1200
=> 70a - \(a^2\) = 1200
=> \(a^2\) - 70a + 1200 = 0
=> \(a^2\) - 30a - 40a + 1200 = 0
=> a(a-30) - 40(a-30) = 0
=>(a-30)*(a-40) = 0 => a = 30 or 40
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Now, we know that b = 140 - 2a and b has to maximum
=> a has to be minimum
=> a = 30
So, b = 140 - 2*30 = 140 - 60 = 80

So, answer will be D
Hope it helps!

perhaps a quicker method would be to Plug-In the Answer Choices, starting from the Largest Value -E- 100 and working down.

Assume the Wall on the 1 Side of the Rectangle = L = Length

The other 3 Sides that make up the Walking Path around the Rectangle will then = 2*Width + Length = 2*W + L = 140

We are Given Also that the Area = L * W = 2,400

Answer Choice E: if L = 100, then W = 24
2*W + L = 148
Since it does not fit our Given Info., we can Eliminate E

Answer Choice D is the Next Largest Value: if L = 80, then W = 30 (30 * 80 = 2,400)
2*W + L = 140
This meets our given info., thus D is the Correct Answer.

Solution:

Let x and y be the dimensions of the playground and let x be the length of the wall. We can create the equations:

xy = 2400

and

x + 2y = 140

So x = 140 - 2y, and substituting this in the first equation, we have:

(140 - 2y) * y = 2400

140y - 2y^2 = 2400

-2y^2 + 140y - 2400 = 0

y^2 - 70y + 1200 = 0

(y - 30)(y - 40) = 0

y = 30 or y = 40

If y = 30, then x = 140 - 2(30) = 80. If y = 40, then x = 140 - 2(40) = 60. Since we are looking for the longest possible length of the wall, we see that it’s 80.

Alternate Solution:

Let x and y be the dimensions of the playground and let x be the length of the wall. We have x + 2y = 140. Since the length of the wall is x, we have length of the wall + 2 * one side of the playground = 140. Using this relation, we can test each answer choice. Let’s start from the greatest answer choice:

E) Length of the wall = 100

If the length of the wall is 100, then 2 * one side of the playground = 140 - 100 = 40. It follows that one side of the playground is y = 20 and the area of the playground is 20 * 100 = 2,000 square meters. This is not consistent with the information given to us.

D) Length of the wall = 80

If the length of the wall is 80, then 2 * one side of the playground = 140 - 80 = 60. It follows that one side of the playground is y = 30 and the area of the playground is 30 * 80 = 2,400 square meters. This agrees with the area of the playground given to us in the question stem.

Answer: D

My thought process was the same, but I thought it should be 2 lengths and 1 width. Why is it the opposite?

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