The angles on the larger base of the trapezoid are 60 degrees, and its circumference is 200cm. How to get maximum area?

Question

The angles on the larger base of the trapezoid are 60 degrees, and its circumference is 200cm. What is the base size of the trapezoid so that its area is the maximum?

This is how far I got. Because of the angles on the base it is obvious we are talking about an isosceles trapezoid which means both sides are the same. I think then we need to get the larger base somehow but I don't see how. After that I am pretty sure what to do next. Make an area formula and then just use the formula for the parabola $\frac{-b}{2a}$ to get x-max and calculate the area. But how do I start this out ? I have no idea how to get any of this with just the information the the trapezoid is isoceles.

Answer 1

you have $$A=\frac{a+c}{2}\cdot h$$ and $a+b+c+d=200$ and $$\sin(60^{\circ})=\frac{h}{b}=\frac{h}{d}$$ can you start from here?

Answer 2

You could calculate things in terms of the height $h$ and the smaller base $b_1$.

Your non-parallel sides $l$ are then each $h/\cos(30^{\circ}) = 2h/\sqrt{3}$, and then your larger base is $b_2 = b_1 + 2 \cdot h/\sqrt{3}$.

Now, take your perimeter $P = b_1 + b_2 + 2l$ and area $A = (b_1 + b_2)h/2$, which upon substituting should each be only in terms of $h$ and $b_1$.

Answer 3

60 degree angles make for special triangles and in this case a special trapezoid.

The perimeter is: $(a+b + 2(a-b)) = 200$

the height is $ (a-b)\frac{\sqrt {3}}{2}$

The area is $\frac {\sqrt {3}}4 (a+b)(a-b)$

To simplify this, lets say:

Let $u = a+b, v = a-b, k = \frac {\sqrt {3}}{4}$

Perimeter $u = 200-2v,$ Area $kuv$

Substituting one into the other give me a function for area in terms of $v$ which will give us a quadratic that we should be able to maximize.

$A = k(v)(200-2v)\\ 2(100kv - v^2)$

Will have a vertex at $v = 50k$

$A = 5000 k^2 = 5000(\frac 3{16}) = 937.5$

And, we have managed to bypass solving for $a$ and $b.$