1. trig problem

Homework Problem:

What is the largest possible area for a parallelogram that has pairs of sides with lengths 5 and 11?

R(subscript)max=?

Thanks

2. Re: trig problem

Originally Posted by jonathanraxa
Homework Problem:

What is the largest possible area for a parallelogram that has pairs of sides with lengths 5 and 11?

R(subscript)max=?

Thanks
The area of the parallelogram is

\displaystyle \displaystyle \begin{align*} A &= a\,b\sin{\theta} \\ &= 5 \cdot 11 \cdot \sin{\theta} \\ &= 55\sin{\theta} \end{align*}

Since this is a sine function that has not been translated vertically, its maximum value is equal to its amplitude.

3. Re: trig problem

Area of parallelogram= base * height Take 11 as the base. Height will be maximum when the parallelogram is a rectangle. So max area=11*5=55

4. Re: trig problem

Originally Posted by biffboy
Area of parallelogram= base * height Take 11 as the base. Height will be maximum when the parallelogram is a rectangle. So max area=11*5=55
Which of course is when \displaystyle \displaystyle \begin{align*} \theta = 90^{\circ} \end{align*} because that maximises the sine function, which I wrote in my post