The length of wire is 1. Let x be the part that is bent into a circle x = 2πr, then 4s = 1-x where s is the square’s side ⇒ s = $\frac{1-x}{4}$. Besides, x = 2πr ⇒ r = $\frac{x}{2π}$. The diagram is shown in Figure 1.a.

$A = π\frac{x^{2}}{4π^{2}}+\frac{(1-x)^{2}}{4^{2}} = \frac{4x^{2}}{4^{2}π}+\frac{π(1-x)^{2}}{4^{2}π} = \frac{4x^{2}+π(1-x)^{2}}{4^{2}π}$

$A = \frac{4x^{2}+π(1-x)^{2}}{4^{2}π} = \frac{4x^{2}+ π -2πx +πx^{2}}{4^{2}π} = \frac{(π+4)x^{2} -2πx + π}{4^{2}π}$

$A’=\frac{1}{4^{2}π}(2(π+4)x -2π)=\frac{1}{8π}((π+4)x -π)$. A’=0 ⇒ x=^{π}⁄_{π+4}

- (0,
^{π}⁄_{π+4}), A’<0 ⇒ A is decreasing. - (
^{π}⁄_{π+4}, 1), A’>0 ⇒ A is increasing. Therefore, x =^{π}⁄_{π+4}is a minimum.

$\frac{s}{r} = \frac{\frac{1-x}{4}}{\frac{x}{2π}} = \frac{(1-x)2π}{4x} = \frac{(1-x)π}{2x}= \frac{(1-\frac{π}{π+4})π}{2\frac{π}{π+4}} = \frac{(\frac{π+4-π}{π+4})π}{2\frac{π}{π+4}}=\frac{1}{2}$

In other words, if **the radius of the circle is half the side of the square, the sum of both areas is the least**.

The length of wire is 1, and therefore their sides are: $\frac{x}{4}, \frac{1-x}{4}$. The diagram is shown in Figure 1.a.

A = $(\frac{x}{4})^{2}+(\frac{1-x}{4})^{2} = \frac{2x^{2}-2x+1}{16}$

A’ = $\frac{4x-2}{16}$. A’=0 ⇒ x=^{1}⁄_{2}.

- (0,
^{1}⁄_{2}), A’<0 ⇒ A is decreasing. - (
^{1}⁄_{2}, 1), A’>0 ⇒ A is increasing. Therefore, x =^{1}⁄_{2}is a minimum. The largest area is in x=0 or 1, that is, a single square (the problem is quite misleading) and its area is^{1}⁄_{16}.

V = x^{2}y = 42,592 ⇒ y = $\frac{V}{x^{2}} = \frac{42,592}{x^{2}}$
The diagram is shown in Figure 1.c.

A = x^{2}+4xy = x^{2}+4^{42,592}⁄_{x} = x^{2}+^{170,368}⁄_{x}

A’ = 2x -^{170,368}⁄_{x2}. A’ = 0 ⇒ 0 = $\frac{2x^{3}-170,368}{x^{2}}$ ⇒ $x = \sqrt[3]{\frac{170,368}{2}}=44$

- (0, 44), A’<0 ⇒ A is decreasing.
- (44, ∞), A’>0 ⇒ A is increasing. x=44 is a minimum ⇒ y = 22 ⇒ Dimensions: 44cm x 44 cm x 22 cm. ⇒ A = 5,808 cm
^{3}.

A’’ = 2 + ^{2*170,368}⁄_{x3} >0 ⇒ Concave upward.

V = x^{2}y (fixed) ⇒ y = $\frac{V}{x^{2}}$

A = x^{2}+4xy = x^{2}+4^{V}⁄_{x}

A’ = 2x -4^{V}⁄_{x2}. A’ = 0 ⇒ 0 = $\frac{2x^{3}-4V}{x^{2}}$ ⇒ $x = \sqrt[3]{2V}$

- (0, $\sqrt[3]{2V}$), A’<0 ⇒ A is decreasing.
- ($\sqrt[3]{2V}$, ∞), A’>0 ⇒ A is increasing. Therefore, x = $\sqrt[3]{2V}$ is a minimum, y = $2^{\frac{-2}{3}}V^{\frac{1}{3}}$

Futhermore, $\frac{x}{y}=\frac{2^{\frac{1}{3}}V^{\frac{1}{3}}}{2^{\frac{-2}{3}}V^{\frac{1}{3}}} = 2.$

Goal: $\frac{ds}{dt}$?

$\frac{dy}{dt}=-60mph, \frac{dx}{dt}=-30mph $.

The diagram is shown in Figure 1.e.

$x^{2} + y^{2}=s^{2} ⇒ 0.6^{2} + 0.8^{2}=s^{2} ⇒ s=1$

$x^{2} + y^{2}=s^{2} ⇒ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=2s\frac{ds}{dt} ⇒ 2·0.6·(-30)+2·0.8·(-60)=2s\frac{ds}{dt} ⇒ \frac{ds}{dt}=-66mph.$

Two triangles are similar when they have equal angles and proportional sides, and therefore $\frac{r}{h} = \frac{4}{10}$ ⇒ r = $\frac{2h}{5}$. The diagram is shown in Figure 1.f.

$V = \frac{1}{3}·base·height=\frac{1}{3}·\pi·r^{2}·h = \frac{4\pi}{75}·h^{3}$

$\frac{dV}{dt}=2$

Goal: h = 5, $\frac{dh}{dt}?$

$\frac{dV}{dt}=2=\frac{4\pi}{75}·3h^{2}·\frac{dh}{dt} = \frac{4\pi}{25}·5^{2}·\frac{dh}{dt} ⇒ \frac{dh}{dt} = \frac{1}{2\pi}$ft/sec.

We know that tan(θ)=^{h}⁄_{2} and $\frac{dθ}{dt} = 5° per second$

Goal: $\frac{dh}{dt}~ when~ θ=50°$. The diagram is shown in Figure 2.c.

$tan(θ)=\frac{h}{2}⇒sec^{2}θ\frac{dθ}{dt} = \frac{1}{2}\frac{dh}{dt}⇒\frac{dh}{dt}=2sec^{2}(50°·\frac{pi}{180})·5°·\frac{pi}{180} ≈ 0.422mps ≈ 1520.7mph$ by 0.422·3.600

All angles should be in radians. To convert degrees to radians, multiply the number of degrees by π/180.

We know that x^{2}+30^{2}=D^{2} (⇒ x = 40) and $\frac{dD}{dt}=-80.$ Goal: $\frac{dx}{dt}?$

The diagram is shown in Figure 2.a.

$2x\frac{dx}{dt} = 2D\frac{dD}{dt}$

Now, we can insert the values in the equation, $2x\frac{dx}{dt} = 2D\frac{dD}{dt} ⇒ 2·40·\frac{dx}{dt} = 2·50·(-80) ⇒ \frac{dx}{dt} = -100 ft/sec.$

|-100 ft/sec| > 95 ft/sec = 65mph. Yes, you are speeding, but not for much.

We know that 3^{2} + b^{2} = c^{2}, $\frac{db}{dt}=480$

480 miles ———- 60*60 seconds

b miles ———— 30 seconds

b = $\frac{480·30}{60·60} = 4 miles$ ⇒ c = 5.

The diagram is shown in Figure 2.b.

2b$\frac{db}{dt} = 2c\frac{dc}{dt}$. Now, we can insert the values in the equation, $2·4·480 = 2·5·\frac{dc}{dt}⇒~ \frac{dc}{dt}=384mph$

It is a root-finding algorithm which produces successively better approximations to the root or zeroes of a real-valued function. The idea is to start with an initial guess, e.g., x_{0} = 2. Then, we approximate the function by its tangent line, and calculate the x-intercept of this tangent line. This x-intercept is our new guess, say x_{1}, and it will typically be a better approximation to the original function's root than the first guess. 2.d.

The tangent line’s formula is: $y-y_0=m(x-x_0)$

x1 is the x-intercept, so $0-y_0=m(x_1-x_0) ⇒ x_1 = x_0 -\frac{y_0}{m} = x_0 -\frac{f(x_0)}{f’(x_0)}$

The process is repeated as: $x_{n+1} = x_n - \frac{f(x_n)}{f’(x_n)}$

Example. Let’s approximate $\sqrt[7]{1000}$. It is the same as trying to find the root of f(x)=x^{7} - 1000.

A reasonably good guess is x_{0}=3. f’(x)=7x^{6}, $x_1 = x_0 -\frac{f(x_0)}{f’(x_0)} = 3 - \frac{3^{7}-1000}{7·3^{6}}≈2.76739173$

$x_2 = x_1 -\frac{f(x_1)}{f’(x_1)} = 2.76739173 - \frac{2.76739173^{7}-1000}{7·2.76739173^{6}}≈2.69008741$

$x_3 = x_2 -\frac{f(x_2)}{f’(x_2)} = 2.69008741 - \frac{2.69008741^{7}-1000}{7·2.69008741^{6}}≈2.68275645$. Google’s search engine returns 2.68269579528 ($\sqrt[7]{1000}$).