Who's good at Calc?

[suede]

Problem 1. A box has a square base and a volume of 1000 cubic
feet. Find the dimensions that minimize the surface area.

Problem 2. A box with a square base is to be constructed using
wood for the sides (at $2 per square foot) and metal for the top
and bottom (at $5 per spare foot). If you have a budget of $100 for
supplies, what's the largest volume the box can hold?

Problem 3. A car accelerates from a stop and then slows to a halt
three minutes later. It's position p relative to the starting line at
time t is given by p = ¡t4 + 18t2, where p is measured in feet and t
in minutes.
(a) How far has the car traveled at the end of the three minutes?
(b) How fast is the car traveling after 1 minute?
(c) When is the car traveling fastest?

[suede]

I think the first one is 15.9 by 15.9 by 3.97....

[suede]

nope scratch that its 10x10x10

[nuttyprofessor]

been a while but I believe you need to establish equations using variables to come up with the answers......or you can rely on the guys on LS1.com for answers.

[suede]

I'm doing it its just making sure im doing it right if soone definately knows what they are doing...

[Eugenio_SS]

problem 1:
square base (x by x) with height h
Volume = hx^2 = 1000, so h = 1000/[x^2]
A = 4 sides + top + bottom
A = 4xh + 2x^2
A = 4x[1000/x^2] + 2x^2
A = 4000/x + 2x^2

minimize de A is when dA/dx = 0
dA/dx = -4000/x^2 + 4x = 0
so 4x = 4000/x^2
so 4x^3 = 4000
so x^3 = 1000
so x = 10... therefore, h = 10 too.

[suede]

see i got that one, same method too...

[Eugenio_SS]

number 2.
similar to problem 1 [square base (x by x) with height h]
V = hx^2

Area(sides) = 4xh @ 2$
Area(bases) = 2x^2 @ 5$
So 100$ = 2$*4xh + 5$*2x^2
100 = 8xh + 10x^2
8xh = -10x^2 + 100
h = 1/4{-5x + 50/x}

since V = hx^2
V = 1/4{-5x + 50/x}x^2
V = 1/4{-5x^3 + 50x}

V is @ max when dV/dx = 0
dV/dx = 1/4{-15x^2 + 50} = 0
15x^2 = 50
x^2 = 10/3
x = SQRT(10/3) so h = (5/6)SQRT(30)

V = hx^2
V = (25/9)SQRT(30)

[Eugenio_SS]

problem 3.
is it a typo: p = ¡t4 + 18t2
Assuming you mean p = it^4 + 18t^2... but what is "i" ??? (typo??)
you sure it's an "i" and not a 1, maybe 8 or 9 (since they are close to "i") ?

[suede]

p=-t^4 + 18t^2

[suede]

btw Eugenio i owe you my life about now... i might make it home tonight yet....

[lemons12]

wow.... i will never talk to Eugenio... he will make me look tarded.... hahahahahaha

i would be by just saying hey
hahahahahhaah

[lemons12]

o btw....

wade your still a bitch for "TSP link

[310stanger]

number 2.
similar to problem 1 [square base (x by x) with height h]
V = hx^2

Area(sides) = 4xh @ 2$
Area(bases) = 2x^2 @ 5$
So 100$ = 2$*4xh + 5$*2x^2
100 = 8xh + 10x^2
8xh = -10x^2 + 100
h = 1/4{-5x + 50/x}

since V = hx^2
V = 1/4{-5x + 50/x}x^2
V = 1/4{-5x^3 + 50x}

V is @ max when dV/dx = 0
dV/dx = 1/4{-15x^2 + 50} = 0
15x^2 = 50
x^2 = 10/3
x = SQRT(10/3) so h = (5/6)SQRT(30)

V = hx^2
V = (25/9)SQRT(30)

somebodies an engineer lol

[suede]

Eugenio, i keep geting something closer to 180 for the volume instead of 15ish for number 2

[LS1Powerhouse]

God I hate calc.. It kicks my ass so much.. I'm studying to be an engineer and damn.. So much math.. So little brain..

[mbrownin]

eugine, i keep geting something closer to 180 for the volume instead of 15ish for number 2

The "15ish" is right, i.e. (25/9)*sqrt(30). I got the same answer. 10x10x10 for the 1st one is correct also.

Car reaches max velocity in SQRT(3) minutes.

[Eugenio_SS]

problem 3:
p=-t^4 + 18t^2, where p is measured in feet and t in minutes.

(a) How far has the car traveled at the end of the three minutes?
t = 3
p(t=3) = -(3)^4 + 18(3)^2
p(t=3) = -81 + 162 = 81ft

(b) How fast is the car traveling after 1 minute?
v = dp/dt
v = -4t^3 + 36t
v(t=1) = -4(1)^3 + 36(1)
v(t=1) = -4 + 36 = 32 ft/min

(c) When is the car traveling fastest?
fastest is when v(t) reaches maximum
ie dv/dt =0
dv/dt = -12t^2 + 36 =0
12t^2 = 36
t^2 = 3
t = SQRT(3) minutes

[Eugenio_SS]

somebodies an engineer lol

Who ???
lucky guess.

[Eugenio_SS]

Eugenio, i keep geting something closer to 180 for the volume instead of 15ish for number 2

I did all my calculations just on top of my head while typing... so I might have made a mental calc error... but by having a second glimpse @ it... the #s still look right.