Maximize the objective $$A = xy,$$ subject to the constraint $$2x+3y = 2400.$$ To find the maximum area, we need to eliminate one of the variables $x, y$ from the objective function. We use the constraint to do this. Solve the constraint for either $x$ or $y$. In this case, solving the constraint for $y$ gives $$y = 800 - \frac 23 x. \;\; \text {(verify)}$$ We substitute this expression of $y$ into the objective function to get $$A(x) = x\left (800 - \frac 23 x\right ) = 800x - \frac 23 x^2.$$ Since $x$ represents a length of fence, we have $x \geq 0$. Furthermore, since we have 2400 feet of fence $2x \leq 2400$ which implies $x\leq 1200$. Thus we would like find the absolute maximum of $A(x)$ on the interval $[0, 1200]$. Since $A(x)$ is a continuous function, the EVT says that it has an abs max on the closed interval $[0, 1200]$. We use the closed interval method of section 3.1 to find the abs max.

The critical numbers are: $$A'(x) = 800 - \frac 43 x = 0$$ which yields $$x= 600 \,\mbox {ft.}$$ which is in the interval. Comparing values, we have $A(0) = 0, A(600) = 240000$ and $A(1200) = 0$. Thus the abs max occurs when $x = 600$.

We can also note that $x = 600$ gives a maximum since the graph of $A(x)$ is a parabola that opens down.

We next find $y$ by plugging $x = 600$ into the constraint: $$y = 800 - \frac 23 x = 800 - \frac 23(600) = 400 \,\mbox {ft.}$$ Finally the maximum area of the enclosure is $$A = xy = (600)(400) = 240{,}000 \,\mbox {ft$^2$}.$$

If we let $x$ represent the length and width of the box, and $y$ its height, then our objective is to maximize the volume,

$$V = \text {length $\times $ width $\times $ height} = x^2y.$$

We have a material constraint which says that the surface area of the box should be 4800 sq. in: $$\text {area of base + area of 4 sides} = x^2 + 4xy = 4800.$$ Solving the constraint for $y$ gives $$y = \frac {4800 - x^2}{4x}.$$ Substituting this into the objective gives: $$V(x) = x^2 \left (\frac {4800 - x^2}{4x}\right ),$$ where $0\leq x \leq \sqrt {4800}$. Simplifying $V(x)$ gives: $$V(x) = 1200x - \frac 14 x^3.$$ The volume will be at a maximum when its derivative is zero: $$V'(x) = 1200 - \frac 34 x^2 = 0.$$ Solving for $x$, we get: $$\frac 34 x^2 = 1200,$$ $$x^2 = 1600,$$ $$x = \pm 40.$$ Since $x=-40$ is not in the interval $[0, \sqrt {4800}]$, we eliminate it from consideration. Furthermore, since $V(0) = 0$ and $V(\sqrt {4800})= 0$ (since $y=0$ in this case), the maximum volume occurs when $x = 40$. Plugging this into the constraint equation, we can find $y$: $$y = \frac {4800 - x^2}{4x} = \frac {4800 - 1600}{160} = 20 \mbox { in.}$$ Thus the volume is maximized when $x = 40$ and $y = 20$ and the maximum volume of the box is: $$V = x^2y = (40)^2 (20) = 32{,}000 \mbox { cubic inches}.$$

To solve this problem, we need the formula for the distance between two points in a plane. If the points are $(x_1, y_1)$ and $(x_2, y_2)$, then the Pythagorean Theorem gives the distance between them: $$d = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$

We apply the distance formula to the comet at the point $(x, y)$ and the sun at the origin to get the objective equation: $$d = \sqrt {(x-0)^2 + (y-0)^2} = \sqrt {x^2 + y^2}.$$ Since the comet must stay on the parabola we get the constraint equation $y = x^2 - 5$. Substituting this into the objective equation gives: $$d = \sqrt {x^2 + (x^2 - 5)^2}.$$ Now, we would like to find the minimum value of $d$. Since the square root function is increasing, we can locate the minimum by finding the min of the radicand (the quantity under the square root). So, we want to minimize $$f(x) = x^2 + (x^2 - 5)^2.$$ The minimum occurs when the derivative is zero, so we solve $f'(x) = 0$ for $x$ yielding $$f'(x) = 2x + 2(x^2 - 5)(2x) = 2x[1+ 2(x^2 - 5)] = 2x(2x^2 - 9) = 0$$ which gives either $$2x = 0 \;\; \text {or} \;\; 2x^2 - 9 = 0.$$ The three solutions are $x = 0, \pm \dfrac {3}{\sqrt 2}$. The corresponding $y$-values can be found by plugging these $x$-values into the parabola $y = x^2 - 5$ giving the three points $$(0, -5), \left (-\dfrac {3}{\sqrt 2}, -\dfrac 12 \right ) \mbox { and } \left (\dfrac {3}{\sqrt 2}, -\dfrac 12 \right ).$$ If the comet is at $(0, -5)$ then its distance to the sun is 5 million miles. If the comet is at either $\left (\pm \dfrac {3}{\sqrt 2}, -\dfrac 12\right )$ then the distance is $$d = \sqrt {x^2 + y^2} = \sqrt {{\frac {9}{2} + \frac 14}} = \frac {\sqrt {19}}{2} \mbox { million miles}.$$ Hence the minimum distance is $\dfrac {\sqrt {19}}{2}$ million miles when the comet is at either of the points $\left (\pm \dfrac {3}{\sqrt 2}, -\dfrac 12 \right )$.

Let $x$ be the length of the enclosure and $y$ the width. Because of the partitions, there are 3 horizontal and 3 vertical sections of fence. Hence the total amount of fence needed can be expressed as $$F = 3x + 3y.$$ This is our objective function. Since the area of the enclosure needs to be 1600 sq. ft., we have the constraint $$xy = 1600.$$ Solving the constraint for $y$ gives $$y = 1600/x.$$ Substituting this into the objective yields $$F(x) = 3x + 3\left (\frac {1600}{x}\right ) = 3x + \frac {4800}{x}.$$ To find the minimum value of $F$ , we solve $F'(x) = 0$. This gives $$F'(x) = 3 - \frac {4800}{x^2} = 0$$ which means $3x^2 = 4800$ and so $x = \pm 40$ ft. Since $x>0$ from context, we have $x = 40$ ft. as our solution. Plugging this value of $x$ into the constraint equation gives $$y = \frac {1600}{x} = \frac {1600}{40} = 40 \mbox { ft.}$$ So the solution is for the gardener to build a square, 40 feet on a side, using a total of $F = 3x+ 3y = 3(40) + 3(40) = 240$ feet of fence.

Let $x$ meters be the distance that the cheetah walks and $y$ meters the distance that the cheetah swims, as shown in the figure below.

The total time for the trip is the time walking plus the time swimming. Since $$\text {time} = \frac {\text {distance}}{\text {rate}},$$ the total time is $$T = \frac {x}{2} + \frac {y}{1}.$$ This is the objective equation. The constraint comes from the Pythagorean Theorem, since the swim portion of the trip is along the hypotenuse of a right triangle with legs parallel and perpendicular to the river bank. One leg of the triangle is $50$ and the other is $100-x$. So $$y = \sqrt {50^2 + (100-x)^2}.$$ Substituting this into the objective gives: $$T(x) = \frac {x}{2} + \sqrt {50^2 + (100-x)^2},$$ where $x$ is between $0$ and $100$. To find the minimum value of $T$, we find the critical numbers (using the chain rule): $$T' = \frac 12 + \frac {2(100-x)(-1)}{2\sqrt {50^2 + (100-x)^2}} = \frac 12 - \frac {100-x}{\sqrt {50^2 + (100-x)^2}} = 0$$

Solving for $x$ gives $$\frac {100 - x}{\sqrt {50^2 + (100-x)^2}} = \frac 12$$ $$2(100 - x) = \sqrt {50^2 + (100-x)^2}.$$ Square both sides: $$4(100-x)^2 = 50^2 + (100-x)^2$$ $$3(100-x)^2 = 50^2$$ $$100-x = \frac {50}{\sqrt {3}} = \frac {50\sqrt {3}}{3}.$$ Hence $$x = 100 - \frac {50\sqrt {3}}{3} \approx 71 \text {meters}.$$ To verify that this is indeed the minimum time, we can use the closed interval method from section 3.2 (Extreme Value Theorem). This is because our independent variable $x$ must be between $0$ and $100$. Thus we compare $T(0), T(100)$ and $T(50/\sqrt 3)$. We have $$T(0) = \frac 02 + \sqrt {50^2 + 100^2} = \sqrt {12500} \approx 111,$$ $$T(100) = \frac {100}{2} + \sqrt {50^2 + 0} = 50 + 50 = 100, \text {and}$$ $$T(50/\sqrt 3) = \frac 12 \cdot \left (100- \frac {50}{\sqrt 3}\right ) + \sqrt {50^2 + (50/\sqrt 3)^2} \approx 93.3$$ Hence the minimum time to reach the prey is approximately $T=93.3$ seconds and this is achieved by walking approximately $x =71$ meters, then swimming the rest of the way.