# Test 3

- Page ID
- 13715

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

Exercise \(\PageIndex{1}\)

A rectangular plot of land is to be fenced in using two kinds of fencing. Two opposite sides will use heavy-duty fencing selling for \(\$3\) a foot, while the remaining two sides will use standard fencing selling for \(\$2\) per foot. What are the dimensions of the rectangular plot of greatest area that can be fenced off at a cost of \(\$6000\)?

**Answer**-
\( 500 ft \times 750 ft\)

**Solution:**-
Let \( x \) and \( y \) be the length and the width of the rectangle fence. Suppose the length is the side cost \(\$3\) a foot, while the width is the side cost \(\$2\) a foot.

Then the cost to produce the fence is \( 6000= 6x+4y\). Thus \(3000=3x+2y\).

We need to maximize the area \(A=xy\).

Since \( 3000= 3x+2y\), \(y=\dfrac{3000-3x}{2}\).

Therefore \( A(x)=x\dfrac{3000-3x}{2}\). Further \(x \in [0,1000]\).

Since the domain is closed and bounded, therefore we will compare the functional values between the endpoints and the critical points.

**Critical points:**Since \( A=x\dfrac{3000-3x}{2}=\dfrac{3000x-3x^2}{2}\), \( A'=\dfrac{3000-6x}{2}\). Therefore \(x= 500\).

x A(x) 0 0 500 (500)(750) 2000 0 We earlier set \(y = \dfrac{3000-3x}{2}\); thus \(y = 750\). Thus our rectangle will have two sides of length 500 and one side of length 750, with a total area of 3750 ft\(^2\).

Exercise \(\PageIndex{2}\)

Verify that the hypotheses of the Mean Value Theorem are satisfied on the given interval and find all values of \(c\) in that interval that satisfy the conclusion of the theorem for \(f(x) = x^2-x \) on \([-3,5].\)

**Answer**-
\(c=1\)

**Solution:**-
Since \(f \) is a polynomial, it is continuous and differentiable everywhere. Therefore, \(f\) satisfies the hypotheses of the Mean Value Theorem, and there must exist at least one value \( c∈(-3,5) \) such that \(f'(c)\) is equal to the slope of the secant line connecting \((-3,f(-3))\) and \((5,f(5)). \)

To determine which value(s) of \(c\) are guaranteed, first calculate the derivative of \(f\). The derivative \(f′(x)= 2x-1\). The slope of the line connecting \((-3,f(-3))\) and \((5,f(5))\) is given by

\[\frac{f(5)−f(-3)}{5+3}=\frac{20−12}{8}=\frac{8}{8}=1.\]

We want to find \(c\) such that \(f′(c)=\frac{7}{4}\). That is, we want to find \(c\) such that

\[2c-1=1\]

Solving this equation for \(c\), we obtain \(c=1\).

Exercise \(\PageIndex{3}\)

Find the position, velocity, speed and acceleration at time \(t=1\), $$ s(t)=9-9\cos\left( \displaystyle \frac{\pi t}{3}\right), 0 \leq t \leq 5.$$

**Answer**-
\(s(1)=4.5 ft. v(1) =1.5 \sqrt{3} \pi, a(1)= \pi^2/2.\)

**Solution:**-
\(s(1)=9-9\cos\left( \displaystyle \frac{\pi }{3}\right)= 9-9(\dfrac{1}{2})=4.5 ft\)

\(v(t)=s'(t)= 3 \pi \sin\left( \displaystyle \frac{\pi t}{3}\right)\), \(v(1)=s'(1)=3 \pi \sin\left( \displaystyle \frac{\pi }{3}\right)= 1.5 \sqrt{3} \pi\)

\(a(t)=v'(t)= \pi ^2 \cos\left( \displaystyle \frac{\pi t}{3}\right)\), \(a(1)=v'(1)=\pi^2 \cos\left( \displaystyle \frac{\pi }{3}\right)=)= \pi^2/2\)

Exercise \(\PageIndex{4}\)

Find each integral

- \(\displaystyle{\int \left( \frac{1}{\sqrt{t}} - 3\sqrt{t} \right) \, dt}\).
- \(\displaystyle{\int 4x\sec^2(x^2)\, dx}\).
- \(\displaystyle{\int \sin^4(x)\cos(x) \, dx}\).
- \(\displaystyle{\int {1 \over 5x+2} \, dx}\).
- \(\displaystyle{\int {\frac{dx}{\sqrt{x}(x+1)}}}\).

**Hint:**-
2. use \(u= x^2\),

3. use \(u= \sin(x) \),

4. use \(u=5 x+2\),

5. use \(u= \sqrt{x}\).

**Answer**-
1. \(2t^{\frac{1}{2}}-2t^{\frac{3}{2}} +C\),

2. \(2 \tan(x^2)+C\),

3. \(\dfrac {\sin^5(x)}{5}+C\),

4. \(\dfrac {\ln(5x+2)}{5}+C\),

5. \(2 \tan^{-1}(\sqrt{x})+C \).

Exercise \(\PageIndex{5}\)

Solve the initial value problem $$ \frac{dy}{dt} = e^{-2t}, y(0)=5.$$

**Answer:**-
\( y= \dfrac {-e^{-2t}+11}{2}\).