Question

a landscape archetiect is hired to build a garden which is of the shape of a right traingle. suppose the architech wants to use a brick wall on the side of AC costing $9/ft and a metal fence on the side costing $3/ft suppose the architect has $1620 budget. what is the maxmum area the architect can build with he budget

Answer 1

Answer:

Maximum Ares architect can build is 36,450 $ft^{2}$

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 = $\sqrt{l^{2} b^{2} }$

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

$\frac{d(area)}{dl}$ = $\frac{\frac{1}{2} [ 540l - 3l^{2}] }{dl}$

0 =  $\frac{1}{2} [ 540 - 6l] }$

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 = $\frac{1}{2}$ x 270 x 270

Maximum Ares = 36,450 $ft^{2}$