# Thread: Max/Min Problems with Trig.

1. ## Max/Min Problems with Trig.

Hi

Could someone please give me hints 4 the questions circled in the attachment? I have no clue how to start to arrive at those answers.

Thanx a lot.

2. Hello xwrathbringerx
Originally Posted by xwrathbringerx
Hi

Could someone please give me hints 4 the questions circled in the attachment? I have no clue how to start to arrive at those answers.

Thanx a lot.
1(a) Let $QR = x$. Then $QP = x\cos\alpha$.

So $x(1+\cos\alpha) = 2$ (= total length of wire)

And $PR = x\sin\alpha$

Then use $A = \tfrac12.PR.QP$

2(a) Draw vertical lines through the top vertices of the trapezium to the base. Then using the right-angled triangles thus formed, the width of the channel across the top is $4+ 4\cos\theta + 4\cos\theta = 4 + 8\cos\theta$, and the depth of the channel $= 4\sin\theta$.

Then use the formula:

Area of a trapezium = $\tfrac12$(sum of parallel sides) $\times$ (perpendicular distance between them)

$=\tfrac12(4 + 4 + 8\cos\theta)\times 4\sin\theta$

$= ...$