A rectangular Problem By Achal Jain

Calculus Level 2

The length of a rectangle varies with time as L = sin ⁡ t L= \sin t and the breadth varies with time as B = cos ⁡ t B= \cos t .

If the time has domain [ 0 , Π 2 ] \left[ 0,\frac { \Pi }{ 2 } \right] , then find the maximum area of the rectangle.


The answer is 0.5.

Achal Jain by Achal Jain , 19, India

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1 solution

Achal Jain
Jul 5, 2016

First we find the rate of change area with respect to time or [ d L B d t ] = d s i n t . c o s t d t = s i n t ( − s i n t ) + c o s t ( c o s t ) \left[ \frac { dLB }{ dt } \right] =\frac { dsint.cost }{ dt } =sint(-sint)+cost(cost) by Chain Rule.

Now the maxima occurs when derivative is zero.

Now we can write the earlier expression as s i n t ( − s i n t ) + c o s t ( c o s t ) = − s i n 2 t + c o s 2 t = c o s 2 t sint(-sint)+cost(cost)=\quad -{ sin }^{ 2 }t\quad +\quad { cos }^{ 2 }t=cos2t .

This is because of trig identity c o s ( a + b ) = c o s a . c o s b − s i n a . s i n b cos(a+b)=\quad cosa.cosb-sina.sinb . So then c o s 2 t = 0 cos2t=0 only when t=90 or Π 4 \frac { \Pi }{ 4 } .

Hence area= s i n t . c o s t sint.cost = 1 2 × 1 2 = 1 2 = . 5 \frac { 1 }{ \sqrt { 2 } } \times \frac { 1 }{ \sqrt { 2 } } =\frac { 1 }{ 2 } =.5

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We could also observe that the area is sin ⁡ ( t ) cos ⁡ ( t ) = 1 2 sin ⁡ ( 2 t ) \sin(t)\cos(t) = \dfrac{1}{2}\sin(2t) , which achieves a maximum of 1 2 \boxed{\dfrac{1}{2}} when sin ⁡ ( 2 t ) = 1 ⟹ t = π 4 \sin(2t) = 1 \Longrightarrow t = \dfrac{\pi}{4} .

Brian Charlesworth - 4 years, 11 months ago

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