A figure is not required to determine the numbers of units to rent to maximize the profit.

The expression to be optimized is given:

The expression to be optimized is given:

and is already a function of a single variable $x$. Observe that $P(x)$ is continuous over the closed interval domain $[0,100]$.

The expression to be optimized is:

Computing $P^{\prime }(x)$:

$P^{\prime }(x)=0$ when $x=1720/40=43$, which is in the open interval $(0,100)$. Therefore, $x=43$ is a critical point.

Since $P(x)$ is continuous on the closed interval $[0,100]$, we could evaluate $P(x)$ at $x=0$, $x=43$, and $x=100$ to determine the number of units that maximizes the profit.

Instead, observe that $P(x)$ is a parabola that opens downward (i.e., $P^{″}(x)=-40$ which means that $P(x)$ is concave down). This implies that the critical point at $x=43$ must correspond to the maximum profit.

Therefore, profit is maximized when 43 units are rented out.

Since we have 200 feet of fencing material, the variables $x$ and $y$ must satisfy the following equation.

Solving for $y$ yields $y=200-2x$. Substituting this into our formula for area from Step 2, we obtain $A(x)$, the area of the region enclosed by the fence as a function of $x$.

Note the following domain restriction: $0<x<100$ since

Start by computing $A^{\prime }(x)$,

which equals zero when $x=200/4=50$.

We can check that $x=50$ corresponds to the maximum area by verifying that the second derivative is negative. Since $A^{″}(x)=-4<0$ for all $x$ in the domain, and $x=50$ is in the domain of $A(x)$, we see that $A(50)$ is the absolute maximum value of $A(x)$ on $(0,100)$.

Solve for $y$ to find the other dimension of the field:

We conclude that the field with maximum area enclosed by 200 feet of fencing material is 50 feet deep and 100 feet wide.

Since the volume of the box is given to be $24$ ft${}^3$, the variables $x$ and $y$ must satisfy the following equation.

Solving for $y$ yields $y=24/x^2$. Substituting this into our formula for cost in Step 2, we obtain $C(x)$, the cost of constructing the box as a function of $x$.

Note the following domain restriction: $x>0$ since each dimension of the box must be positive.

Start by computing $C^{\prime }(x)$,

which equals zero when $x^3-8=0$ (i.e., $x=2$).

Finally, we can check that the critical point $x=2$ corresponds to the minimum cost by verifying that the second derivative is positive. Since

is positive for all $x$ in the domain (i.e., $x>0$), and $x=2$ is in the domain of $C(x)$, we see that $C(2)$ is the absolute minimum value of $C(x)$ on $(0,\mathrm{\infty })$.

We can now solve for $y$, and write down the dimensions of the box.

Therefore, the dimensions of the box that minimize the cost of construction are $2\times 2\times 6$.

A figure is not required to determine the numbers of units required to maximize the profit.

The total profit function $P(x)$ is given by:

The total profit function $P(x)$ is given by:

Note the following domain restriction: $0<x<1250$ since

Start by computing $P^{\prime }(x)$:

which is equal to zero when $x=2440/4=610$. Note that $x=610$ is in the domain of $P(x)$.

We can confirm that $x=610$ corresponds to the maximum profit by verifying that the second derivative is negative. Since $P^{″}(x)=-4<0$ for all $x$ in the domain, and $x=610$ is in the domain, we see that $P(610)$ is the absolute maximum profit.

Therefore, in order to maximize profits, the company must produce and sell 610 units (at a price of $p(610)=2500-2\cdot 610=1280$ dollars per unit).