TIFR CSE 2016 | Part A | Question: 9

Question

Suppose a rectangular farm has area $100$ square meters. The lengths of its sides are not known. It is known, however, that all the edges are at least $2$ meters in length. Which of the following statements about the rectangle's perimeter $p$ (in meters) is FALSE?

• A. $p$ can take all values between $45$ and $50$
• B. $p$ can be $52$ for some configuration
• C. $p$ can take all values between $55$ and $60$
• D. $p$ can be $70$ for some configuration
• E. $p$ can be $39$ for some configuration

Comments

Formula for perimeter of rectangle is 2(x+y)

where x and y are length and breadth of the rectangle.

So perimeter of rectangle is always even.(multiplied by 2)

Using this property we can easily  find out that 39 is false since it's a odd.

Answer 1

perimeter cant be 39.
min possible value of length and breadth with l>2 and b>2 which satisfy area l*b = 100 is length=10 and breadth =10
with this min value perimeter = 2(l+b) = 2(10+10) = 40
min possible value of perimeter = 40
so perimeter cant be 39

Comments

my solution is wrong but i think uptil here it is correct P= 48 - (lb)/2 now one more condition is b = 100/l+2 -2  and we will get a function then we have to find the max and min i think...as they did not mention l and b are integer values

Answer 2

Let length of farm = $x$ meter

      Width of farm = $y$ meter

Area = $xy$ =100

Parameter = $P$ = $2(x+y)$ = $2(x+\frac{100}{x})$

$\frac{\mathrm{d} p}{\mathrm{d} x}$ = $2(1-\frac{100}{x^2})$ =$0$

So $x$ = 10 Meter // using Maxima/minima concept we can see that's point of minima for perimeter

So minimum Perimeter =$2(10+10)$=$40$

Graph would be kind of Hyperbolic in +ve xy axis so continuous mean every parameter values that is greater than or equal to 40 is possible.

Answer 3

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Comments

you meant max value of perimeter possible is 48?