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, a 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 $



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