Afarmer wishes to enclose two identical adjoining rectangular pens against the side of a large barn. she wants each pen to have an area of 650 square feet650 square feet and will use the barn as one side of the enclosure. what is the least amount of fencing needed (in feet) to create the two pens?

L(min)   =  125  ft

Dimensions of the pens

biggest side x /2 =  31.225  ft

shorter   side  w =  20.82  ft

Step-by-step explanation:

Let call   w   width  of the pen    ( the shorter side )

Let call   x    the biggest side of the two pen.

The configuration we get is that we will use 3 times w  +  just one side of x

Then total length of fence is

L  =  3*w  + x    (1)

And area of one pen  is    A = (x/2)* w     and that area have to be 650 ft²

650*2 = x*w         ⇒    w  =  1300/x

So to get L as function of  x we plugg this value in equation (1)

L(x)   =  3* 1300/x   +  x

Taking derivatives on both sides of the equation

L´(x)   =  -3900/x²  +  1

L´(x)  =  0         ⇒     x²   =  3900      ⇒  x  =  62.45  ft

and   w  =  1300/x        w  =  20.82 ft

And the least amount of fence is

L(min)   =  3* 20.82   +   62,45

L(min)   =  62.45  + 62.45

L(min)   =  125  ft

explanation:

remove parenthesis:

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multiply fractions:

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divide:

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mordancy.

step-by-step explanation:

there's the chart