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

#### MarkFL

Staff member
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.

#### MarkFL

Staff member
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:

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