Hello Kendra Leota,
1.) I would first draw a sketch of the corrals:
View attachment 1204
We can now see that we have the objective function (the enclosed area $A$):
$$A(x,y)=xy$$
subject to the constraint (the amount of fencing $F$):
$$2x+3y=F$$
I know of 3 methods to use here. The first two will require that we express $A$ as a function of 1 variable, so if we solve the constraint for $y$, we find:
$$y=\frac{F-2x}{3}$$
and by substitution we now have:
$$A(x)=x\left(\frac{F-2x}{3} \right)=\frac{1}{3}\left(x(F-2x) \right)$$
i) Pre-calculus technique:
We see that we have a parabolic function opening down, and so the vertex will be the global maximum. The vertex must be midway between the roots, and we can see the roots are at:
$$x=0,\,\frac{F}{2}$$
and so the axis of symmetry, which contains the vertex, is the line:
$$x=\frac{F}{4}$$
and thus:
$$y=\frac{F-2\left(\frac{F}{4} \right)}{3}=\frac{F}{6}$$
We then see that half of the fencing is in the horizontal segments and half is in the vertical segments.
ii) Using differentiation
Let's write the area function as:
$$A(x)=\frac{1}{3}\left(Fx-2x^2 \right)$$
Now, equate the first derivative to zero to find the critical point:
$$A'(x)=\frac{F}{3}-\frac{4}{3}x=0\implies x=\frac{F}{4}$$
$y$ is found in the same way as the first method. We see that:
$$A''(x)=-\frac{4}{3}<0$$
and so we know the function is concave down everywhere, so our critical value is a maximum.
iii) Multi-variable technique (Lagrange multipliers):
We obtain the following system:
$$y=2\lambda$$
$$x=3\lambda$$
This implies:
$$2x=3y$$
and putting this into the constraint, we find:
$$x=\frac{F}{4},\,y=\frac{F}{6}$$
To finish up, letting $$F=200\text{ ft}$$, we find the dimensions which maximize the enclosed area is:
$$x=50\text{ ft}$$
$$y=\frac{100}{3}\text{ ft}$$
2.) Let's draw a sketch:
View attachment 1205
We can see that we have the area of the rectangle as:
$$A(x,y)=2xy$$
Using the fact that $y$ is a function of $x$, we find:
$$A(x)=2x\sqrt{r^2-x^2}$$
Using the given $$r=8$$, we have:
$$A(x)=2x\sqrt{8^2-x^2}$$
To find the value of $x$ when $A=30$, we may write:
$$30=2x\sqrt{8^2-x^2}$$
Square both sides:
$$900=4x^2\left(64-x^2 \right)$$
$$225=x^2\left(64-x^2 \right)=64x^2-x^4$$
$$x^4-64x^2+225=0$$
Observing that we have a quadratic in $x^2$, we may apply the quadratic formula to obtain:
$$x^2=\frac{64\pm\sqrt{(-64)^2-4(1)(225)}}{2(1)}=32\pm\sqrt{799}$$
Hence, taking the positive root as we require $0\le x\le8$):
$$x=\sqrt{32\pm\sqrt{799}}$$
3.) Again, let's draw a sketch:
View attachment 1206
We now see that we may relate the two known sides of the triangle to the angle using the tangent function:
$$\tan(\theta)=\frac{x}{d}\implies x=d\tan(\theta)$$
Using the given $$d=300\text{ ft}$$ we have:
$$x(\theta)=300\tan(\theta)\text{ ft}$$
To find the distance traveled by the rocket when $\theta=20^{\circ}$, we may write:
$$x\left(20^{\circ} \right)=300\tan\left(20^{\circ} \right)\text{ ft}\approx109.191070279861\text{ ft}$$
