Applications Of Derivatives Question 43
Question: A wire 34 cm long is to be bent in the form of a quadrilateral of which each angle is $ 90{}^\circ $ . What is the maximum area which can be enclosed inside the quadrilateral-
Options:
A) $ 68cm^{2} $
B) $ 70cm^{2} $
C) $ 71.25cm^{2} $
D) $ 72.25cm^{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let one side of quadrilateral be x and another side be y so, $ 2(x+y)=34 $ or, $ (x+y)=17….(i) $
We know from the basic principle that for a given perimeter square has the maximum area, so, x = y and putting this value in equation (i) $ x=y=\frac{17}{2} $
Area $ =x.y=\frac{17}{2}\times \frac{17}{2}=\frac{289}{4}=72.25 $
BETA
