4.7 Applied optimization problems  (Page 5/8)

 Page 5 / 8

Modify the area function $A$ if the rectangle is to be inscribed in the unit circle ${x}^{2}+{y}^{2}=1.$ What is the domain of consideration?

$A\left(x\right)=4x\sqrt{1-{x}^{2}}.$ The domain of consideration is $\left[0,1\right].$

Solving optimization problems when the interval is not closed or is unbounded

In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value theorem, we were guaranteed that the functions had absolute extrema. Let’s now consider functions for which the domain is neither closed nor bounded.

Many functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For example, the function $f\left(x\right)={x}^{2}+4$ over $\left(\text{−}\infty ,\infty \right)$ has an absolute minimum of $4$ at $x=0.$ Therefore, we can still consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema. In the next example, we try to minimize a function over an unbounded domain. We will see that, although the domain of consideration is $\left(0,\infty \right),$ the function has an absolute minimum.

In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has the smallest surface area.

Minimizing surface area

A rectangular box with a square base, an open top, and a volume of $216$ in. 3 is to be constructed. What should the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?

Step 1: Draw a rectangular box and introduce the variable $x$ to represent the length of each side of the square base; let $y$ represent the height of the box. Let $S$ denote the surface area of the open-top box.

Step 2: We need to minimize the surface area. Therefore, we need to minimize $S.$

Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base. The area of each of the four vertical sides is $x·y.$ The area of the base is ${x}^{2}.$ Therefore, the surface area of the box is

$S=4xy+{x}^{2}.$

Step 4: Since the volume of this box is ${x}^{2}y$ and the volume is given as $216\phantom{\rule{0.2em}{0ex}}\text{in}{.}^{3},$ the constraint equation is

${x}^{2}y=216.$

Solving the constraint equation for $y,$ we have $y=\frac{216}{{x}^{2}}.$ Therefore, we can write the surface area as a function of $x$ only:

$S\left(x\right)=4x\left(\frac{216}{{x}^{2}}\right)+{x}^{2}.$

Therefore, $S\left(x\right)=\frac{864}{x}+{x}^{2}.$

Step 5: Since we are requiring that ${x}^{2}y=216,$ we cannot have $x=0.$ Therefore, we need $x>0.$ On the other hand, $x$ is allowed to have any positive value. Note that as $x$ becomes large, the height of the box $y$ becomes correspondingly small so that ${x}^{2}y=216.$ Similarly, as $x$ becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval $\left(0,\infty \right).$ Note that, unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval. However, in the next step, we discover why this function must have an absolute minimum over the interval $\left(0,\infty \right).$

Step 6: Note that as $x\to {0}^{+},$ $S\left(x\right)\to \infty .$ Also, as $x\to \infty ,$ $S\left(x\right)\to \infty .$ Since $S$ is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some $x\in \left(0,\infty \right).$ This minimum must occur at a critical point of $S.$ The derivative is

${S}^{\prime }\left(x\right)=-\frac{864}{{x}^{2}}+2x.$

Therefore, ${S}^{\prime }\left(x\right)=0$ when $2x=\frac{864}{{x}^{2}}.$ Solving this equation for $x,$ we obtain ${x}^{3}=432,$ so $x=\sqrt[3]{432}=6\sqrt[3]{2}.$ Since this is the only critical point of $S,$ the absolute minimum must occur at $x=6\sqrt[3]{2}$ (see [link] ). When $x=6\sqrt[3]{2},$ $y=\frac{216}{{\left(6\sqrt[3]{2}\right)}^{2}}=3\sqrt[3]{2}\phantom{\rule{0.2em}{0ex}}\text{in}.$ Therefore, the dimensions of the box should be $x=6\sqrt[3]{2}\phantom{\rule{0.2em}{0ex}}\text{in}.$ and $y=3\sqrt[3]{2}\phantom{\rule{0.2em}{0ex}}\text{in}.$ With these dimensions, the surface area is

$S\left(6\sqrt[3]{2}\right)=\frac{864}{6\sqrt[3]{2}}+{\left(6\sqrt[3]{2}\right)}^{2}=108\sqrt[3]{4}\phantom{\rule{0.2em}{0ex}}\text{in}{.}^{2}$

application of function
how i can need help
what ?
Bunyim
defination of math
azam
application of function
azam
azam
what is a circle
Ronnie
show that the f^n f(x)=|x-1| is not differentiable at x=1.
is there any solution manual to calculuse 1 for Gilbert Strang ?
I am beginner
Abdul
I am a beginner
ephraim
what is mathematics
logical usage of numbers
Leo
thanks
Henry
you welcome
Leo
what's career can one specialize in by doing pure maths
Lucy
Lucy Omollo...... The World is Yours by specializing in pure math. Analytics, Financial engineering ,programming, education, combinatorial mathematics, Game Theory. your skill-set will be like water a necessary element of survival.
David
***onetonline.org/find/descriptor/result/1.A.1.c.1
Bruce
mathematics seems to be anthropocentric deductive reasoning and a little high order logic. I only say this because I can only find two things going on which is infinitely smaller than 0 and anything over 1
David
More comprehensive list here: ***onetonline.org/find/descriptor/result/1.A.1.c.1?a=1
Bruce
so how can we differentiate inductive reasoning and deductive reasoning
Henry
thanks very much Mr David
Henry
hi everyone
Sabir
is there anyone who can guide me in learning the mathematics easily
Sabir
Hi Sabir first step of learning mathematics is by falling in love with it and secondly, watch videos on simple algebra then read and solved problems on it
Leo
yes sabir just do every time practice that is the solution
Henry
it will be work over to you ,u know how mind work ,it prossed the information easily when u are practising regularly
Henry
in calculas,does a self inverse function exist
Lucy
I'm lost in all functions need help
Jonathan
hello i need help in rate of change
Moises
***questioncove.com/invite/QzOQGp
Bruce
hello my name is Charles Christian
Hello Charles
Jianna
Hi! I am Dante
Dante
Hi! I'm ashwini
ashwini
halĺo
Roben
Hi
Leo
hello leo
Agboke
can anyone prove why AU(BnC)=(AUB)n(AUC)
Agboke
this one it can't be proven these are assumption
Henry
hello agboke there is no proof for such
Leo
Hi
hi this is wasim
wasim
can anybody put me through flowchart and algorithm here
Agboke
Leo
Luis
music while you math
Luis
dy/dx= 1-cos4x/sin4x
what is the derivatives of 1-cos4x/sin4x
Alma
what is the derivate of Sec2x
Johar
d/dx(sec(2 x)) = 2 tan(2 x) sec(2 x)
AYAN
who knows more about mathematical induction?
Agboke
who know anything about the whole calculus thing 🤔 its killing me 😶
matbakh
Yes
What is the integral of 2xe^x
Okay, so the chat works when I use the browser but why does the chat not work in the app?
Bruce
what is lim x=0 Tan 5x/x
integration of 3 cos^2 x - 1
keerthi
Test
Bruce
hillo
Naseer
3/4 sin2x+ 1/2 x+ C
differentiate f(x)=1/2√x
1/-2x^2
chrispine
some symbols are not shown property
Mr
f(x)=1/2âx
Mr
how to solve e for slope eequations
I am willinv to know this. Though i know little bit.
m = y2 -y1 / x2 - x1
Reymund
explain mercuries theorem?
limit of f(x) sec x - 1 /1 - sec x while x is approaching 0
hi is someone in there
hello
This
hi
Agboke
please can anyone recommend a very good maths text book for me? 100 level student
Agboke
for algebra?
Elysha
Hi
Gajji
I don't know all these as i just got in
Agboke
I trying to remember haven't done this in years
Chavon
Paul
okay thanks
Agboke
hi
Rabbit
yes
Nelson
hi sir
Rabbit
Hello
hassan
Beta function
Allu
what is xsin (1/x)
1
The denominator of a fraction is 5 more than the numerator. If half the numerator plus one is added to both terms of the fractions, the resulting fraction would be 5/6. Find the original fraction.
Conney
what is (x+1/x)
Agboke