Trig Problems!!! (1 Viewer)

[survivor]

please anyone...
i have this assignment to do on trig but i cant do a couple of the questions if anyone could help it would be much appreciated

1.) Find the min and max values of 2secx - tanx for 0<(or equal to) x <(or equal to) pie/4

2.) A canned fruit producer wishes to minimize the area of sheet metal used in manufacturing cans of a given volume. Find the ratio of radius to height for the desired can . (Can is a circular cylinder with closed ends)

thankyou in advanced to anyone who can help me out

 

[kewpid]

1)
<pre>
let y =2secx-tanx
=(2/cosx)-(sinx/cosx)
=(2-sinx)/cosx
dy/dx=[(cosx)(-cosx)-(2-sinx)(-sinx)]/(cosx)^2
=2sinx-1 = 0 (for stationary points)
therefore x = pi/6

test nature (blahdiblah do this yourelf, you will find it is a local minimum)
x=pi/6, y= sqrt(3)

since the function is continuous in the domain 0<=x<=pi/4 , therefore (pi/6,sqrt(3)) is an absolute minimum turning point.

thus, to find maximum value of x, we must test the endpoints (ie,x=0 and x=pi/4)

when x=0, y=2
when x=pi/4, y=2srt(2)-1=1.82


therfore max value = 2
therefore minimum value = sqrt(3)
</pre>

2) i can't work it out, not enough info i think

 

[wogboy]

let V = volume of can, A = surface area of can, h = height, r = radius.

V = pi*(r^2)*h ...(A)

and

A = 2*pi*r^2 + 2*pi*r*h ... (B)
(This is the area in terms of two variables. In order to be able to minimise A, we need to get A in terms of only ONE variable, and then differentiate, i.e. we need to eliminate either h or r)

we can change the top equation in terms of h,

h = V/(pi*r^2)

so,

A = 2*pi*r^2 + 2*pi*r*V/(pi*r^2)
A = 2*pi*r^2 + 2*V/r

(note that V is constant, since the question says "cans of a given volume")

so we have one variable A, in terms of one other variable r. Now the question asks us to minimise A, so now can differentiate:

dA/dr = 4*pi*r - 2*V/r^2

setting dA/dr = 0 (to find minumum A),

4*pi*r - 2*V/r^2 = 0
4*pi*r = 2V/r^2
r^3 = V/(2*pi)

now sub in V from the very top equation (A),

r^3 = pi*(r^2)*h/(2*pi)
r^3 = (r^2)*h/2
r = h/2

therefore,

r/h = 1/2 (this is the ratio of radius to height)

So the height of the can is to be double its radius (or equal its diameter), for the can to have minumum surface area for a given volume.

 

[survivor]

thankyou so much both of you