Marc L. answered • 11/18/20

Scored an 800 on SAT Math section

So this is an optimization problem (a personal favorite of mine), we need 2 equations to solve for what we are looking for, in this case a perimeter and an area, so lets make those:

Perimeter) 2L+2W=500 but one side doesn't need fencing since it is next to a barn, so the actual equation relating to amount of fencing used should be the following: L+2W=500

Area=LW

now lets solve for L in the perimeter equation and plug into the area equation:

L=500-2W ---> Area=W(500-2W)=500W-2W^{2}

Now we have our area equation in terms of w, now lets take the derivative and set it =0 to solve for maximum area:

Area'=500-4W=0, 500=4W, **W=125**

Now we have w, lets plug it back into l to solve for the length:

L=500-2(125), **L=500-250=250**

So the Dimensions of the fence should be **(LxW) 250x125**