Question

If three sides of trapezoid are each 6 inches long, how long must the fourth side be if the area is a maximum.

Answer 1

To illustrate the problem, it is better to draw the figure as follows

Since the altitude of the trapezoid is not given in the problem, then label further the above figure as follows

Apply Pythagorean Theorem at the two right triangles of a trapezoid in order to get the value of h, we have

$c^2=a^2+b^2;$

$6^2=h^2+x^2;$

$36=h^2+x^2;$

$h^2=36-x^2;$

$h=\sqrt{36-x^2};$

We know that the area of a trapezoid is

$A=\frac12 \times h\times (b_1+b_2);$

Substitute the values of h, b₁, and b₂ to the above equation, we have

$A=\frac12 \times \sqrt{36-x^2} \times (6+2\times x+6);$

$A=\sqrt{36-x^2} \times (x+6);$

Take the derivative on both sides of the equation with respect to x, we have

$\frac{dA}{dx}=\frac{d}{dx}[\sqrt{36-x^2} \times (x+6)];$

$\frac{dA}{dx}=\sqrt{36-x^2} \ \frac{d}{dx} (x+6) + (x+6) \ \frac{d}{dx} \sqrt{36-x^2};$

$\frac{dA}{dx}=\sqrt{36-x^2} - \frac{x^2+6 x}{\sqrt{36-x^2}};$

Set $\frac {dA}{dx}=0$ because we want to maximize the area of a trapezoid

$0=\sqrt{36-x^2} - \frac{x^2+6 x}{\sqrt{36-x^2}};$

$\sqrt{36-x^2} = \frac{x^2+6 x}{\sqrt{36-x^2}};$

$36-x^2=x^2+6x;$

$2x^2+6x-36=0;$

$(x+6)(x-3)=0;$

Equate each factor to zero and solve for the value of x:

If

x+6=0;

x=-6;

Since the value of x is negative, then it is not accepted as a part of a length of a base of a trapezoid.

If

x-3=0;

x=3;

Since the value of x is positive, then it is accepted as a part of a length of a base of a trapezoid.

Therefore, the length of the fourth side of a trapezoid is

$b_2=2x+6; \\ b_2=12 \ (in);$

Result: 12 inches.