Question:

A rectangular field is to be fenced on three sides, with the fourth side being a straight river. The goal is to maximize the area of the field using 100 meters of fencing. Determine the dimensions of the field that will result in the maximum area.

Answer:

Let's assume the width of the rectangular field to be $w$ meters and the length be $l$ meters. We are given that only three sides of the field will be fenced and the fourth side is along a straight river. This means that only $w + 2l$ meters of fencing can be used.

We need to find the dimensions of the field that will result in the maximum area. The area of a rectangle is given by $A = lw$. However, we have a constraint that the total fencing available is 100 meters, so we can express this constraint mathematically as $w + 2l = 100$.

To find the dimensions that maximize the area of the field, we need to express the area $A$ in terms of a single variable. We can do this by solving the constraint equation for one of the variables. Let's solve for $w$:

Substituting this value of $w$ in the equation for area:

We have now expressed the area of the field, $A$, as a function of one variable $l$. To find the dimensions that maximize the area, we need to find the critical points of this function and determine whether they correspond to maximum or minimum values.

Let's find the critical points by taking the derivative of $A$ with respect to $l$ and setting it equal to zero:

Solving this equation for $l$:

Now, we need to determine whether this critical point corresponds to a maximum or minimum. We can use the second derivative test for this. Taking the second derivative of $A$ with respect to $l$:

Since the second derivative is negative, this implies that the critical point $l = 25$ corresponds to a maximum.

Now that we have determined the value of $l = 25$, let's substitute it back into the constraint equation to find the width $w$:

Therefore, the dimensions of the field that will result in the maximum area using 100 meters of fencing are a width of 50 meters and a length of 25 meters.

To summarize:

• Width of the field, $w = 50 \, \text{meters}$
• Length of the field, $l = 25 \, \text{meters}$

The maximum area of the field can be found by substituting these values into the area formula: