Maximizing the area of an isosceles trapezoid

[MarkFL]

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]

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:

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

subject to the constraint:

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

hence:

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

Using Lagrange multipliers, we find:

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

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

which implies:

$$4h^2=x^2-16^2$$

Substituting this into the constraint yields:

$$x^2-16x-128=0$$

Discarding the negative root, we find:

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