Arancher with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle.
(a) find a function that models the total area of the four pens.
(b) find the largest possible total area of the four pens.

a) A(r) = ( 1/2) * (750*x - 5*x²)

b)  Dimensions

x  =  75 ft

y  =  187,5 ft

A (max) = 14062,5 ft²

Step-by-step explanation:

Fencing material available   750 ft

Rectangula area  A(r)

Let  x  and  y dimensions of rectangular area, and x the small side of the rectangle, then

The perimeter of the rectangle is

P(r)  = 2*x  + 2*y     (1)

To get the four pens we have to place three more fence in between the two x sides of the rectangle, in such way that the total fence is

P(r) + 3*x  = 750

So       2*x + 2*y + 3*x   =  750      ⇒  5*x  + 2*y  =  750   ⇒  y = (750- 5x)/2

Plugging that value in (1)

A(r)  =  x* y

A(r) = x* ( 750 - 5*x)/ 2    ⇒  A(r) = ( 1/2) * (750*x - 5*x²)

Taking derivatives in both sides of the equation we get:

A´(r)  = (1/2) * ( 750 - 5*x)      ⇒  A´(r)  = 0    ⇒     (1/2) * ( 750 - 10*x)  = 0

750 - 10*x  = 0     ⇒  x =  750/10     ⇒   x  =  75 ft

And  y would be

y  = (750- 5x)/2      ⇒   y  =  (750 - 5*75) / 2    ⇒  y  =  375 / 2

y  =  187,5 ft

A(max)  =  187,5*75

A(max)  = 14062,5 ft²

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

you can't solve x because it is 2 linear equation and you only have one equation