Welcome to our community

Be a part of something great, join today!

Cris' question at Yahoo! Answers regarding maximizing the area of an isosceles trapezoid

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is the question:

An isosceles trapezoid has a lower base of 16 cm and the sloping sides are each 8 cm.?
find the width of the upper base for greatest area. application of maxima and minima.
I have posted a link there to this topic so the OP can see my work.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Cris,

We know the lower base is 16 and we may let the upper base be $x$. Thus, the objective function, the area of the trapezoid is:

\(\displaystyle A(h,x)=\frac{h}{2}(16+x)\)

subject to the constraint:

\(\displaystyle \left(8-\frac{x}{2} \right)^2+h^2=8^2\)

hence:

\(\displaystyle g(h,x)=x^2-32x+4h^2=0\)

Using Lagrange multipliers, we find:

\(\displaystyle \frac{16+x}{2}=\lambda(8h)\)

\(\displaystyle \frac{h}{2}=\lambda(2x-32)\)

which implies:

\(\displaystyle 4h^2=x^2-16^2\)

Substituting this into the constraint yields:

\(\displaystyle x^2-16x-128=0\)

Discarding the negative root, we find:

\(\displaystyle x=8\left(1+\sqrt{3} \right)\)