<p>Hello, a seemingly easy calc AB problem has been bothering me for quite some time and I was wondering if any of the self-proclaimed math "geeks" could show me how this problem is done. I am currently studying applications of derivatives and more specifically modeling and optimization. Anyways, I had to recreate the problem with a picture. I appreciate any responses!</p>
<p>okay this is pretty simple you first need an equation modelling the troughs area:
i guess it would be the area of the trapezoid (its in your book!) * 20
now you just find the deriv of x</p>
<p>Since the third dimension (the 20 ft) is constant, you only need to deal with the other 2 dimensions. The equation of the area of a trapezoid is (1/2)(b+B)(h), where b is the base of the trapezoid, and B is the other base (forget the actual terms). Since you know b=1, solving for B in terms of x gives 2<em>sqrt(1-x^2)+1 (do pythagorean theorem on the two triangles formed by hypotenuse 1 and leg x, and add 1). Next, you form the equation for area as a function of x, which would be f(x)=(1/2)(2</em>sqrt(1-x^2)+1+1)(x), which simplifies to f(x)=x(sqrt(1-x^2)+1). Take the derivative of this, set it equal to zero and find the x value. Remember also to take the second derivative to make sure that this value is a max.</p>
<p>Thanks guys, I really just have trouble coming up with the equations. I came up with an answer, but I am still unsure if its the correct solution.</p>
<p>v= 1/2(a+b)h *20
I came up with..
v= 1/2(1+(1+2sinx))(cosx)(20)
v= 1/2(2+2sinx)(cosx)(20)
v=(1+sinx)(cosx)(20)
v=(cosx+sinxcosx)(20)</p>
<p>Then i just graphed..max value of x is .7680
Can anyone confirm this?</p>
<p>EDIT- Sorry tompumpkin did not see your post. I will try it your way (and most likely the correct way).</p>
<p>TowerPumpkin has the right idea, but x is stated as an angle, not the length of a leg. You can solve for x = leg<em>length and then back</em>calculate to find the value of the angle.
Alternatively, if you use x= the angle shown, the area of the trapezoid works out to 0.5(b+B) h, where b=1, B = 1+2sinx, h= cosx. The area A simplifies to
A = (1 + sinx) cosx</p>
<p>dA/dx works out to -sinx + (1 - 2sinx.sinx) . Set t = sinx, and solve the quadratic
-2t^2 -t + 1 = 0; the possible values are t= -1 or t=0.5 . Setting sinx = t= -1 would give you an area A=0, which clearly won't maximize anything :-). The answer is sinx = t= 0.5, or x = 30 degrees. The area is A = (1 + 0.5) ( sqrt(3)/2) .</p>
<p>I was on the right track. Thanks optimizer! Math has never been a problem in years past, but it just seems that this year during calc, I have not been able to learn the material as easy. Some of the problem lies in my pre-requisite math skills, but they are starting to come back during exercises like these.</p>
