The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is
- 90000
- 160000
- 120000
- 100000
The Correct Option is C
Approach Solution - 1
To solve the problem of finding the lowest possible cost of fencing the rectangular plot, we need to work with both the perimeter for the cost and the area to satisfy the given condition.
Let the length of the plot be l and the breadth be b. The area of the rectangle is given as:
l×b=60000
The perimeter cost differs along different sides. Let's enumerate the fencing cost:
- Cost per foot of one side (length), l: ₹ 200
- Cost per foot of the other three sides (2 breadths and 1 length), 2×b+l: ₹ 100
The total cost for fencing would then be:
Cost=200×l+100×(2×b+l)
Simplifying,
Cost=200l+100×(2b+l)=200l+200b+ 100l=300l+200b
The problem requires minimizing this cost while having lb=60000.
From the area equation:
b=60000l
Substitute b in the cost equation:
Cost=300l+200×60000l
Simplifying:
Cost=300l+12000000l
To find minimum cost, take derivative with respect to l:
dCostdl=300−12000000l²
Set derivative to zero for minimum cost:
l³=40000
Solving, we get l=200 and subsequently b=60000200=300.
Substituting back, we compute minimal cost:
Cost=(300×200×+200×300)=120000
Thus, the minimum cost of fencing all sides is ₹ 120000.
Approach Solution -2
Let the length and breadth of the region be represented by \(x\) and \(y\) respectively.
Then the area of the region is given by:
\(xy = 60000\) \(\dots \, (i)\)
The cost of fencing the region is calculated by:
\(200x + 100x + 100y + 100y\)
This simplifies to:
\(300x + 200y\)
Now, we know from the Arithmetic Mean-Geometric Mean inequality (AM-GM inequality) that:
\(\frac{300x + 200y}{2} \geq \sqrt{300x \times 200y}\)
Thus, we get:
\(300x + 200y \geq 2\sqrt{300x \times 200y}\)
Substituting the value of \(xy\) from equation \((i)\):
\(300x + 200y \geq 2\sqrt{300 \times 200 \times xy}\)
Substitute \(xy = 60000\) into the equation:
\(300x + 200y \geq 2\sqrt{300 \times 200 \times 60000}\)
This simplifies to:
\(300x + 200y \geq 2\sqrt{3600000000}\)
\(300x + 200y \geq 2 \times 60000\)
Finally, we get:
\(300x + 200y \geq 120000\)
Thus, we can conclude that the cost will always be greater than or equal to:
\(120000\)
Therefore, the correct option is (C): \(120000\).
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