MAXIMUM AREA OF A PEN PROBLEM

MAXIMUM AREA OF A PEN PROBLEM

(c.f. Grove HSC 2U EX 2.9/2.10, Grove HSC 3U EX 2.10/2.11)

The farmer's wife, Hazel, got $180$ m of fencing at a sale. Hazel asked her husband, Bob, if he would build a rectangular fence of the largest possible area. Hazel wants to know the dimensions of the rectangle and its area.


Background. Two rectangles can have the same perimeter but have different areas. For example;

The area of the left is much bigger than the right but they both have the same perimeter!

To solve this problem Bob let's the sides of the rectangle be length $x$ and height $y$ as shown.

Then Perimeter $P=2x+2y$. Bob is given the constraint that perimeter $P=180$ m.

Therefore

$ 2x+2y=180$

$\therefore 2y=180-2x $

$\therefore y=90-x $

Area of Pen is $A=xy $

$\therefore A=x(90-x) $
$$\therefore A=90x-x^2 $$

Bob  wishes to find the value of $x$ which gives a maximum value for  $A$.

'$A$' is a function of $x$.

Bob  uses calculus to solve this problem.

Differentiating with respect to $x$ gives (twice),

$${dA\over dx}=90-2x$$

$${d^2A\over dx^2}=-2$$

[REMARK: ${dA\over dx}$ is the same thing as $A'$. In problems involving maximum and minimum values it is more common to use ${dA\over dx}$ to make clear what we are differentiating with respect to. ]

For stationary points: $A'=0$.

$\therefore 90-2x=0$

$x=45$.

When $x=45$, $A=90x-x^2=90(45)-45^2=2025$.

Stationary point is $(45,2025)$.

When $x=45$, ${d^2A\over dx^2}=-2<0$. So concave down.

This means that $x=45$ is a maximum turning point.

Bob tells his wife that the rectangle should be of dimensions $45$  m $\times 45$ m to maximise the area (i.e. a square shape). 

The area of the maximum rectangle is $2025$ cm${}^2$.

GROVE HSC 2U EX 2.9 Q10 - not a stationary point, with asymptotes.

This function goes up to $\infty$ and down to $-\infty$ so it has no maximum or minimum value. (Grove's answer is wrong)

STATIONARY POINTS EXAMPLE WITH AN INFLEXION AS WELL

QUESTION: Consider the curve $$y=x^3-3x^2-9x+27$$
(a) Find the coordinates of any stationary points and determine their nature.
(b) Find the coordinates of any points of inflexion (and verify using a y'' box that concavity changes)
(c) Get the $x$ and $y$ intercepts. (its possible here)
(d) Sketch it!
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Before we begin just find y' and y'' right away, factorise both of them ready for use later!

$y'=3x^2-6x-9$
  $y'=3(x^2-2x-3)$
  $y'=3(x-3)(x+1)$                                                (1)

$y''=6x-6$
$y''=6(x-1)$                                                          (2)


(a) For Stat Pts: solve $y'=0$.

 $3(x-3)(x+1)=0$                                      from (1)

$\therefore x=-1, 3 $

For $x=-1, y=(-1)^3-3(-1)^2-9(-1)+27=32$

For $x=3, y=(3)^3-3(3)^2-9(3)+27=0$

Stationary Points are  $(-1, 32)$  and  $(3,0)$.
Now determine their nature using a $y'$ box for each point.

For $(-1, 32)$.

 For $(3, 0)$.

[REMARK: Instead of a $y'$ box we could use $y''$ to test concavity at each $x$ value, to determine the nature. This works only if $y''\neq 0$.
Since $y''(-1)<0$ it's concave down at $x=-1$, so it's a MAX.
Since $y''(3)>0$ it's concave up at $x=3$, so it's a MIN. ]

(b)  For possible inflexion, solve $y''=0$

$6(x-1)=0$, 
$\therefore x=1$.

When $x=1$, $y=(1)^3-3(1)^2-9(1)+27=16$
Possible Inflexion is $(1,16)$.

Verify that is an inflexion using $y''$ box.

(c)  For $y$ intercept put $x=0$.  $\therefore y=27$ For $x$ intercept put $y=0$.  $\therefore 0=x^3-3x^2-9x+27$. By inspection we see that $x=-3,3$.

[REMARK: its often the case that you can't easily find the $x$ intercepts. Only do this as necessary or unless asked to do so in the question!]

(d) 

GROVE HSC 2U EX 2.6 Q11 STATIONARY POINT - TRICKY FUNCTION

QUESTION: Find any stationary points on the curve $y=(x-3)\sqrt{4-x}$

This is a product of two functions, so we must use the Product rule to differentiate it.

If $y=uv$, $$y'=vu'+uv'$$

$u=x-3$         $u'=1$

$v=\sqrt{4-x}$
$v=(4-x)^{1\over 2}$
$v'={1\over 2}(4-x)^{-{1\over 2}}(-1)$
$v'=\displaystyle {-1\over 2\sqrt{4-x}}$

[RECALL chain rule. If $y=f(x)^n$, then $$y'=n\,f(x)^{n-1}\times f'(x)$$]

Therefore,
$y'=vu'+uv'$
$y'=\sqrt{4-x} .1 + (x-3)\left( \displaystyle {-1\over 2\sqrt{4-x}}\right)$

$y'=\displaystyle {\sqrt{4-x} \over 1} + \displaystyle{-(x-3) \over 2\sqrt{4-x}}$

$y'=\displaystyle {\sqrt{4-x} \over 1} + \displaystyle{3-x \over 2\sqrt{4-x}}$

$y'=\displaystyle {\sqrt{4-x} \over 1}\times {2\sqrt{4-x} \over 2\sqrt{4-x}} + \displaystyle{3-x \over 2\sqrt{4-x}}$

[RECALL: $\sqrt{a}\times \sqrt{a}=a$]

$y'=\displaystyle{2(4-x)+3-x \over 2\sqrt{4-x}}$

$y'=\displaystyle{8-2x+3-x \over 2\sqrt{4-x}}$

$y'=\displaystyle{11-3x\over 2\sqrt{4-x}}$

FOR STAT POINTS: SOLVE $y'=0$.

$\displaystyle{11-3x\over 2\sqrt{4-x}}=0$

$\therefore 11-3x=0$

$3x=11$

$x=\displaystyle {11\over 3}$

When
$x=\displaystyle {11\over 3}$,
$y=\displaystyle\left({11\over 3}-3\right)\sqrt{4-{11\over 3}}$
$y=\displaystyle{2\over 3}\sqrt{{1\over 3}}$
$y=\displaystyle{2\over 3\sqrt{3}}$

$y=\displaystyle{2\sqrt{3}\over 9}$

Stationary point is $\displaystyle\left({11\over 3},{2\sqrt{3}\over 9}\right)$.

Now determine the nature of this stat point.

So it's a maximum turning point.

Correct to 2 decimal places it's $(3.67, 0.38)$. (Grove gives the answer to 2 d.p. even though the question didn't ask for it)

GROVE HSC 2U EX 2.6 Q5 POINT OF INFLEXION

QUESTION: Show that the curve $f(x)=2x^3-5$ has an inflexion and find its coordinates.

$f'(x)=6x^2$
$f''(x)=12x$

For possible inflexions: solve $f''(x)=0$.

$12x=0$
$\therefore x=0$.
When $x=0, f(0)=2(0)^3-5=-5$.

So, possible inflexion is $(0,-5)$.

Check concavity to verify inflexion.

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PS. the graph wasn't asked for but here it is anyway :-)

GROVE HSC 2U EX 2.6 Q1 - Stationary Point (Min)

Thx Holly :-)

GROVE HSC 2U EX 2.6 Q13 - stat point problem

GROVE HSC 2U EX 2.6 Q14 - stat points problem

GROVE HSC 2U EX 2.6 Q15 - PROBLEM SOLVING USING INFLEXION PROPERTY.

GROVE HSC 2U EX 2.7 Q14 with horizontal/vertical asymptote

Thx Ellie & Stephy

GROVE HSC 2U EX 2.6 Q8 stats points example using y''