Expert: Scotto - 2/5/2010

Question
Can you please help me with this problems?
1. A rectangular building will be constructed on a lot in the form of a right triangle with legs of 60ft. and 80ft. If the building has one side along the hypotenuse, find its dimensions for maximum floor area.

2. The material for the top and sides of a cylindrical can costs twice as much per square unit as that for the bottom. Find the proportion for least cost.

3. Find the area of the largest isosceles triangle inscribed in a semicircle of radius 10ft. the vertex of the triangle being at the center of the circle.

plzzzz help me...
thx...

Answer
1. The lot is 60 by 80, so the diagonal has length L = √(60ē+80ē) = √(3600+6400) = √10,000 = 100.
Making the hypotenuse the x axis with the corner of the building on the y axis gives us the
equations for the two sides.  This makes the triangle have sides of length 3*20, 4*20, and 5*20.

Since the big triangle is similar the the triangle in equach of the quadrants of our graph,
it can be seen the that base of the left triangle is 3*12 with height 4*12 and hypoteneuse 5*12.
The triangle on the right side will have base 4*16, height 3*15, and diagonal 5*16.

That is, the left triangle will have sides 36(base), 48 (height), and 60 and
the right triangle will have sides 64 (base), 48 (height), and 100.

It should be noted that both triangles will have height 48.
The bases wil be 36 and 64, and 36 + 64 = 100

Draw a picture.  That makes the words a whole lot clearer.

The diagonal line as the edge of the lot on the left side of the y axis has the equation
y = 4x/3 + 60.  The diagonal line on the right as the equation y = -3x/4 + 48.
Note that in both equations, the y value has to be the same, but the x values will be different,
so lets make the 2nd equation into y = 4z/3 + 60, where z is also on the x axis.

The area will be A = y(x-z).
Solve y = 4z/3 + 60 for z in terms of y and put that in the equation.
Put in y = -3x/4 + 48 in place of y in the area equation.

Now that x is the only variable, multiply it out, take the derivative in terms of x,
set it to , and solve for x.


2. Let r be the radius of the cyllinder and h be the height.
The area of the top of a cylinder is πrē.  The sides have area 2πrh.
The area of the bottom is the same as the area of the top.
The total cost is then πrē + 2πrh  + 2πrē = π(rē + 2rh + 2rē) = π(3rē + 2rh).

Let V be the volume of the can that is needed.  Then we have V = πrēh, so h = V/(πrē).
Taking that expression for h and puttining it in π(3rē + 2rh) will gives us an equation in r
for the total cost.

Once this has been done, take the derivative (noting V is a constant),
set it equatl to 0, solve for r, use the V equation to find h.

The proportion will be r/h.


3. This means the one endpoint of that triangle is at the center of a 10ft circle.
We need to choose the angle T to maximize the area of the triangle.

Given angle T and the radius r=10, cut the angle in half.
We know the sides of the triangle are A=r*sin(T/2) and B=r*cos(T/2).
Now that area is AB/2, but there are two triangles that are reflective of each other.
This makes the total area AB.  Note that AB = rēsin(T/2)cos(T/2).

Since 2*sin(A)cos(A) = sin(2A), it can be said that sin(T/2)cos(T/2) = sin(T)/2.
This makes the area equation that same as f(T) = rēsin(T)/2, where f is for area.

Maximizing means setting f'(T) = 0.
Since f'(T) = rēcos(T)/2, setting the to0 will give cos(T) = 0, and that is at T = 90°.

This means the triangle of the largest area will be the one where the central angle between
the sides is 90°.