2 Calculus questions for you to play with?

Question

Call me stupid but Im jumping ahead of the class and want to get these 2 questions done before tomorow hopefully.

1) An open rectangular concrete water tank is to have a volume of 62.5 cubic meters. Determine the least surface area required.

2) A container in the shape of a right circular cone of height 225mm and base radius of 22.5mm is catching the drips from a tap leaking at the rate of 1100 mm cubed / second. Find the rate at which the surface area of water is increasing when the water is half way up the cone.

Workings out are needed so I can understand what youve done.

Im sure its probably really easy and I'l look stupid but Im learning still! Any help much appreciated, thanks

Answer 1

Matt, it's anything but stupid to try to learn MORE! Don't say that, say "Call me overly ambitious..."

Anyway, I assume you're in Calc 1, so for the first one, you'd have to know if the base is a square, otherwise there are 3 unknown dimensions. We can do it with 2 dimensions (assuming base is a square with length x) as follows:

V = x²h
62.5 = x²h
h = 62.5/x²

A = x² + 4xh
A = x² + 4x(62.5/x²)
A = x² + 250x^-1
A' = 2x - 250x^-2
0 = 2x - 250x^-2
270x^-2 = 2x
125 = x^3
x = 5, therefore h = 62.5/25 = 2.5

2) This is a related rates exercise in which dV/dt (the change in volume over time) = 1000 mm^3/s. You are asked to find dh/dt when h = 112.5 mm. In a cone, the radius and height of the volume of the WATER is proportional to the original dimensions of the conical container (think similar triangles). So the ratio of height to radius is 10 to 1, right? Thus, h/r = 10/1 --> .1h = r. We substitute that into the volume formula for a cone:

V = πr²h/3
V = π(.1h)²(h)/3
V = πh³/300
You are asked to find dh/dt when h = 112.5 mm. Now differentiate with respect to t:

dV/dt = πh²/100 dh/dt
1100 = π(112.5)²/100 dh/dt
dh/dt = 110000/112.5²π mm/s
dh/dt = 704/81π mm/s

Answer 2

1. The rectangular shape with minimum surface is a cube.
Therefore the size of your cube will be third root of 62.5 = 3.97 m and the surface will be 6*3.97^2 = 94.5 m^2.

2. The volume of a right circular cone of radius r and height h is:
v = pi*h*r^2/3 = s*h/3..........(1),
where h is the height, s is the base surface, and r is the base radius. I presume cone upside down meaning the base is up. It doesn't affect the result
Also in your case
h/r = 225/22.5 = 10...........(2).
From
s = pi*r^2 = pi*h^2/100........(3)
follows: h = sqrt(100*S/pi) and from (1)
v = (10/(3*sqrt(pi))*(s^(3/2)).
From there it follows:
dv/dt = (10/(3*sqrt(pi))*(3/2)*sqrt(s)*ds/dt, and with (3)
= (10/2)*r*ds/dt, and with (2)
= (5/10)*h*ds/dt, which gives:
ds/dt = (4/225)*dv/dt = 4*1100/225 = 19.55555556
mm^2/sec.

Answer 3

I´ll show you the first
The problem is
V=xyz given=62.5
S= xy +2xz+2yz minimum
Using Lagrangian multipliers we have
W =xy +2xz+2yz+k(V-xyz) minimum
We have ro put the partial derivatives of W to 0
Wx=y+2z-kyz=0 (a)
Wy=x+2z-kxz=0 (b)
Wz=2x+2y-kxy=0 (c)
V=xyz(d)
Multiplying (a)by x and (b)by y and substracting we get
2z(x-y)=0 and as z can´t be 0 x=y
From (c) we get
x(4-kx)=0 and as x can´t be 0 k=4/x
From (b) x+2x-4z=0 so x=2z
At last we have
x=2z y=2z and V = 4z^3 =62.5m^3
so z= 2.5m x=5m and y=5m
I´ll let you prove that it is a real minimum,and I don´t think that´s really easy