
# Test 3

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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 $$\6000$$?

$$500 ft \times 750 ft$$

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 $$[-3,5].$$

$$c=1$$

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 $$t=1$$, $$s(t)=9-9\cos\left( \displaystyle \frac{\pi t}{3}\right), 0 \leq t \leq 5.$$

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

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

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

2. use $$u= x^2$$,

3. use $$u= \sin(x)$$,

4. use $$u=5 x+2$$,

5.  use $$u= \sqrt{x}$$.

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

$$y= \dfrac {-e^{-2t}+11}{2}$$.