MHB Maximize Area of Isosceles Trapezoid: Solve w/ Maxima & Minima

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To maximize the area of an isosceles trapezoid with a lower base of 16 cm and sloping sides of 8 cm, the upper base width must be determined. The area function is defined as A(h,x) = (h/2)(16+x), with a constraint involving the height h. By applying Lagrange multipliers, the relationship between the variables is established, leading to the equation x^2 - 16x - 128 = 0. Solving this yields the optimal upper base width of x = 8(1 + √3). This approach effectively demonstrates the application of maxima and minima in geometric optimization.
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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.
 
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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:

$$x=8\left(1+\sqrt{3} \right)$$
 
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