Suppose a farmer encloses a rectangular region of a land next to a river. fencing will be used on 3 sieds, and none is needed along the river. tje farmer ha 180 feet of fencing available to use. using x and y for the dimensions of the rectangle. the equation for the amount of fencing used on each side is 2x u

Dimensions :

x (the longer side, only one side with fence )   =  90  ft

y ( the shorter side two sides with fence )        =  45  ft

Total fence used   45 * 2  +  90    =  180 ft

A(max)  =  

Step-by-step explanation: If a farmer has 180 ft of fencing to encloses a rectangular area with fence in three sides and the river on one side, the farmer surely wants to have a maximum enclosed area.

Lets call "x" one the longer side  ( only one of the longer side of the rectangle will have fence, the other will be along the river and won´t need fence. "y" will be the shorter side

Then we have:

P = perimeter  =  180  =  2y  +  x        ⇒   y  =  ( 180 - x )  / 2        (1)

And   A (r)   =  x * y

A(x)   =  x  *  ( 180 - x ) /2       ⇒   A(x)   = (180/2) *x   -  x² / 2

Taking derivatives on both sides of the equation :

A´(x)   =  90   -  x    

Then if    A´(x)   =  0   ⇒        90   -  x    =  0     ⇒   x  =  90 ft

and from :       y  =  ( 180 - x )  / 2     ⇒     y  =  90/2

y  =  45  ft

And

A(max)  =  90 * 45    =  4050  ft²

1 1/7 hope i

step-by-step explanation:

8/7 = (1 * 7 + 1) / 7 = 1 + 1/7 = 1 1/7

c = 0.429t

step-by-step explanation:

the constant rate at which carlos harvests cassava is 35 minutes per 15 cassavas 15/35 = 0.429 cassava per minute

therefore, an equation that models the number of cassava (c) that harvest carlos in a time (t) is the following:

where t is the time in minutes.

this means that when 35 minutes have elapsed 15 cassava have been harvested, at 70 minutes 30 cassava have been harvested and so on.