Question 1159033
<pre>
{{{drawing(400,1600/9,-1,11,-1,5,

line(0,0,10,0),
line(10,4,10,0),
line(0,4,10,4),
line(0,0,0,4),
line(5,0,5,4),

locate(0,4.5,A), locate(5,4.5,B), locate(10,4.5,C),
locate(0,0,F), locate(5,0,E), locate(10,0,D),


locate(2.5,0,x),locate(7.5,0,x),
locate(2.5,4.5,x),locate(7.5,4.5,x),
locate(-.3,2,h),locate(4.7,2,h),locate(9.7,2,h) )}}}

Cost = $20∙(AB+BC+CD+DE+EF+FA) + $50∙(BE)

Cost = 20∙(x+x+h+x+x+h) + 50∙(h)

Cost = 20(4x+2h) + 50h

Cost = 80x+40h+50h

Cost = 80x+90h

Area = AC∙CD = (2x)∙(h) = 1600

2xh = 1600

h = 1600/(2x) = 800/x

Cost = C = 80x+90(800/x)

{{{matrix(3,3,

C,""="",80x+72000x^(-1),
dC/dx,""="",80-72000x^(-2),
dC/dx,""="",80-72000/x^2)}}}

Set the derivative equal to 0, to find potential extrema:

{{{matrix(8,3,

80-72000/x^2,""="",0,
80x^2-72000,""="",0,
80x^2-72000,""="",0,
80(x^2-900),""="",0,
x^2-900,""="",0,
x^2,""="",900,
x,""="","" +- sqrt(900),
x,""="","" +- 30)}}}

Now we test to see if x=30 produces the minimum cost.
We first try the 2nd derivative test

{{{matrix(4,3,

dC/dx,""="",80-72000/x^2,
dC/dx,""="",80-72000x^(-2),

(d^2C)/(dx^2),""="",0+144000x^(-3),
(d^2C)/(dx^2),""="",144000/x^3)}}}

When we substitute 30 for x, we will get a positive number,
so the curve is concave upward at x=30, so x=30 produces
the minimum cost.

So the dimensions are when x=30 ft. and h = 800/x = 800/30 = 26 2/3 ft.
But since x is only half the base, the actual dimensions are 60 ft by 26 2/3 ft.

The minimum cost is Cost = C = 80(30)+7200/30 = $2640.

Edwin</pre>