Answer:
Maximum Ares architect can build is 36,450 [tex]ft^{2}[/tex]
Explanation:
Suppose
The Perpendicular of the triangle = l
The Base of the triangle = b
The hypotenuse of the triangle = h
The hypotenuse of the triangle can be calculated as follow
h = [tex]\sqrt{l^{2} b^{2} }[/tex]
Total Budget = $1,620
As the architect does not want to build anything on the hypotenuse of the triangle area.
So as per given condition
( $9 x l ) + ( $3 x b ) = $1,620
$3 ( 3l + b ) = $1,620
3l + b = $1,620 / $3
3l + b = 540
b = 540 - 3l ...........(1)
Area = 1/2 x l x b
using (1) we will have
Area = 1/2 x l x ($540 - 3l)
Now differentiating w.r.t l
[tex]\frac{d(area)}{dl}[/tex] = [tex]\frac{\frac{1}{2} [ 540l - 3l^{2}] }{dl}[/tex]
0 = [tex]\frac{1}{2} [ 540 - 6l] }[/tex]
0 = 540 - 6l
6l = 540
l = 540/6
l = 90 ft
Placing value of l in (1)
b = 540 - 3(90)
b = 540 - 270
b = 270
So, Maximum area will be calculated as follow
Maximum Ares = [tex]\frac{1}{2}[/tex] x 270 x 270
Maximum Ares = 36,450 [tex]ft^{2}[/tex]
