Arectangular area of 36 f t2 is to be fenced off. three sides will use fencing costing $1 per foot and the remaining side will use fencing costing $3 per foot. find the dimensions of the rectangle of least cost. make sure to use a careful calculus argument, including the argument that the dimensions you find do in fact result in the least cost (i. e. minimizes the cost function).

x = 8,49 ft

y  = 4,24  ft  

Step-by-step explanation:

Let  x be the longer side of rectangle   and  y  the shorter

Area of rectangle     =    36 ft²     36  =  x* y   ⇒ y =36/x

Perimeter of rectangle:

P  =  2x   +   2y    for convinience we will write it as    P  = ( 2x + y ) + y

C(x,y)   =  1 * ( 2x  +  y  )  +  3* y

The cost equation as function of x is:

C(x)  =  2x  + 36/x   + 108/x

C(x)  =  2x  + 144/x

Taking derivatives on both sides of the equation

C´(x)  = 2  - 144/x²

C´(x)  = 0         2  - 144/x² = 0       ⇒  2x²  -144 = 0    ⇒  x² =  72

x = 8,49 ft       y  = 36/8.49    y  = 4,24  ft    

How can we be sure that value will give us a minimun

We get second derivative

C´(x)  = 2  - 144/x²      ⇒C´´(x)  = 2x (144)/ x⁴

so C´´(x) > 0

condition for a minimum

an altitude of 15,000 feet

step-by-step explanation:

250

step-by-step explanation: