We need to enclose a rectangular field with a fence. We have 300 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area

If we are looking to enclose the largest area possible with 300 feet of fencing material, then we need to look at the most efficient way in which this can be accomplished. The best approach is to use a rectangular shape, as it will offer us the most flexibility in working with our given length of fencing material.

The first step is to determine how much fencing we have available for each side of our rectangle. Since one side has already been determined by the building, we must split our remaining 300 ft between three sides of the rectangle. Thus, each side should measure 100 ft in length. This also means that all four sides of our rectangle will be equal lengths since they are all being measured by the same value (100ft).

Now that we know how long each side needs to be, let’s explore how their widths will affect the total area enclosed by the fence. To do this, let’s set up an equation using basic geometry: A = L * W where A represents Area and L stands for Length while W stands for Width. We know that L = 100ft due to our previous calculations; however, W is still unknown at this point and can thus become a variable when substituting into this equation as such: A = 100ft * W where W now stands for some unknown width value which results from algebraic manipulation on both sides of said equation and leaves us with A/100ft = W or more simply put – Width = Area/Length (in other words: what is needed inside a box divided by what’s outside).

We need to enclose a rectangular field with a fence. We have 300 feet of fencing material and a building is on one side of the field and so won’t need any fencing. Determine the dimensions of the field that will enclose the largest area

Since we have 300 ft of fencing material available total and no information about its thickness or weight capacity per linear foot (we could assume something thin like chain-link but would not want to overestimate), let’s assume that any excess above 400 sq ft enclosed would add too much strain on whatever type of fence might be used downscale from large farm fences–which tend towards heavier materials like wood planks or steel posts–and thus reduce its longevity when compared against scenarios involving fewer square feet within a smaller boxed-off space instead (for example 350 sq ft versus 500 sq ft). This information helps narrow down potential solutions significantly because it now gives us an upper limit on maximum area allowable before risking structural integrity issues with whatever supporting posts might otherwise hold said fence so safely together if larger sizes were chosen without due care taken concerning their overall strength requirements as well as resiliency against weather conditions over time etc… Using these assumptions then leads us towards one optimal solution here being:
Area ≤ 400sq ft ⟹W ≤ 4(L) And since L=100ft —> Width should take on values less than or equal too 4*100=400ft
Additionally note that since these dimensions represent only two sides out of four required ones needed make up one closed off enclosure around our rectangular shaped field—they must also adhere strictly too following rule simultaneously as well if desired outcome wishes remain fulfilled successfully : i) 200ft + 400ft >300ft ; ii) 200+400<600 . Now plugging 400fts back into original equation yields results giving maximum possible area achievable through use herein assuming aforementioned constraints get respected accordingly meaning: A=(L)(W)=(100)*(400)=40,000sq  feet                                           Amax=40,000sq feet                                      Amax≤400sq feet Note though once again depending upon exact desired application final design may need modified slightly further based upon individual preference accessed via trial/error experimentation process until satisfactory resolution found.

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