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Had a couple problems. Stuck on the second. Just wanted to check my work and that I had my theory correct:

  1. Find the dimensions of a rectangle with area $1000\mathrm{m}^2$ whose perimeter is as small as possible.

Answer

So,

$$L \cdot w = 1000\mathrm{m}^2$$ $$w = \frac{A}{L} = \frac{1000\mathrm{m}^2}{L}$$

Perimeter is = $ 2L + 2(\frac{1000\mathrm{m}^2}{L})$

So

$$P(L) = 2L + 2 \cdot \frac{1000}{L}$$ $$P'L = 2 + 2 \cdot \frac{-1000}{L^2}$$

the derivative $= 0$ when:

$$-1 = \frac{-1000}{L^2}$$ $$L^2 = 1000$$ $$L = 10 \sqrt{10}$$

and corresponding $W$ since $A = 1000$ is $\frac{100}{\sqrt{10}}$

Does this look right?

  1. Consider the following problem: A farmer with $750$ ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?

(a) Draw several diagrams to illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. Find the total areas of these configurations. Does it appear that there is a maximum area? If so, estimate it.

(b) Draw a diagram illustrating the general situation. Introduce notation and label the diagram with your symbols.

(c) Write an expression for the total area.

(d) Use the given information to write an equation that relates the variables.

(e) Use part (d) to write the total area as a function of one variable.

(f) Finish solving the problem and compare the answer with your estimate in part (a).

Answer How are (d) and (e) different btw?

a) It looks like a square pen with sides of $125$ft and $3$ parallel rows of $125$ each leads to an Area of $15625$. This is the largest $A$ I can create by hand. So there are $5$ total sides (the outside and the $3$ interior paralell) of $125$ft each.

b) I have a box with 3 parallel sides. So I think my perimeter equation is:

$$2x + 5y = 750$$

c) $x \cdot y = A$

but also:

d) $$2x + 5y = 750$$ $$5y = \frac{750}{2x}$$ $$y = \frac{750}{10x}$$

e) $$x \cdot \frac{750}{10x} = A$$ $$A' = \frac{-7500X}{100x^2} + \frac{750}{10x}$$

Where do I go from here?

f)

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    $\begingroup$ $2x + 5y = 750 \implies y = \frac {750 - 2x}{5}$ $\endgroup$ – Doug M Mar 9 '18 at 20:49
  • $\begingroup$ Rationalise the denominator of $W$ in question $1$, and you may notice something interesting. $\endgroup$ – daviewales Mar 9 '18 at 21:13
  • $\begingroup$ What is the diff between (d) and (e) $\endgroup$ – Jwan622 Mar 9 '18 at 21:19
  • $\begingroup$ YOOOOOOOO @daviewales $\endgroup$ – Jwan622 Mar 9 '18 at 21:20
  • $\begingroup$ Perhaps d can be any equation which correctly relates the variables, but e must be rearranged for area? $\endgroup$ – daviewales Mar 9 '18 at 21:27
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The first problem solution is right.

The second problem formulation is right. However: $$2x+5y=750 \Rightarrow 5y=750-2x \Rightarrow y=150-0.4x.$$ I think you can handle the rest.

Addendum: Once you get one variable area function $A=x(150-0.4x)$, you will equate its derivative to zero like you did in the first problem: $$A'=150-0.8x=0 \Rightarrow x=187.5 \Rightarrow y=150-0.4\cdot 187.5=75.$$

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  • $\begingroup$ what is the diff between (d) and (e) $\endgroup$ – Jwan622 Mar 9 '18 at 21:19
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$$2x + 5y = 750$$

and

$$5y = \frac{750}{2x}$$

are not equivalent.

You wanted $$ 5y=750-2x$$ instead.

.

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