Thread: Applied minima/maxima etc

I have a quiz tomorrow and im having a little bit of problem understanding how to work with these questions. I understand fine Dirivatives / etc but not these sepcific problems.

1) An open-topped box having square horizontal cross section is to hold 108 cubic Feet of grain when full. What demensions should it have if its total area (base plus four vertical sides) is to be as small as possible.

so a) we're looking for the area being minimized
and b) Im not sure what the function would be.

after i find the function, i know im supposed to:
differentiate the function and find the zeroes of the first derivative to identify the stationary points which are potential minimum or maximum and then use the second derivative test to classify each stationary point as min / max


2) Find the least area of a triangle in the first quadrant formed by the coordinate axes in a line through the point (3,8)

and 3) A box is to be made from a 10 inch by 16 inch sheet of metal by cutting equal squares out of the corners and bending up the flaps to form sides. Find the dimensions that maximize the volume

This time i know the vunction is V= x(10-2x)(16-2x) .. But i am not sure why thats the function basically.



If you have time to help, id very much appreciate it!
Ty!

Originally Posted by 3deltat

I have a quiz tomorrow and im having a little bit of problem understanding how to work with these questions. I understand fine Dirivatives / etc but not these sepcific problems.

1) An open-topped box having square horizontal cross section is to hold 108 cubic Feet of grain when full. What demensions should it have if its total area (base plus four vertical sides) is to be as small as possible.

so a) we're looking for the area being minimized
and b) Im not sure what the function would be.

after i find the function, i know im supposed to:
differentiate the function and find the zeroes of the first derivative to identify the stationary points which are potential minimum or maximum and then use the second derivative test to classify each stationary point as min / max


2) Find the least area of a triangle in the first quadrant formed by the coordinate axes in a line through the point (3,8)

and 3) A box is to be made from a 10 inch by 16 inch sheet of metal by cutting equal squares out of the corners and bending up the flaps to form sides. Find the dimensions that maximize the volume

This time i know the vunction is V= x(10-2x)(16-2x) .. But i am not sure why thats the function basically.



If you have time to help, id very much appreciate it!
Ty!

Here are similar problems to questions 1 and 3. can you figure out what to do from these?

http://www.mathhelpforum.com/math-he...-open-box.html

http://www.mathhelpforum.com/math-he...ents-help.html

http://www.mathhelpforum.com/math-he...lving-box.html

http://www.mathhelpforum.com/math-he...-problem3.html

http://www.mathhelpforum.com/math-he...-problem2.html

http://www.mathhelpforum.com/math-he...n-problem.html

mm yes im able to see most of the concept from thoes problems

However for V= x(10-2x)(16-2x) specifically, why is it -2x?

and im not quite certain how to get the Function for the first two. I can get the rest after the function though.

Originally Posted by 3deltat

3) A box is to be made from a 10 inch by 16 inch sheet of metal by cutting equal squares out of the corners and bending up the flaps to form sides. Find the dimensions that maximize the volume

always draw a diagram for problems like this. see the diagram below.

let x be the length of the sides of the squares that are cut out from the corners. so if we consider each side, we will realize that a total length of 2x has been removed (see the diagram), that's where the -2x comes from.

so we have the base of the box has length (16 - 2x) and width (10 - 2x) and height x.

now V = length*width*height = x(16 - 2x)(10 - 2x), so we have:

V = x(160 - 32x - 20x + 4x^2)
=> V = 160x - 52x^2 + 4x^3
=> V' = 160 - 104x + 12x^2

for max and min, set V' = 0

=> 12x^2 - 104x + 160 = 0
=> 3x^2 - 26x + 40 = 0
=> (3x - 20)(x - 2) = 0
=> x = 20/3 oe x = 2
now we find which of these is a minimum using the second derivative formula:

V'' = -104 + 24x
V''(20/3) = -104 + 24(20/3) = 56 > 0 ....this is the local min
V''(2) = -104 + 24(2) = -56 < 0 ...........this is the local max

so the dimensions of the box that maximizes the volume are:
height = 2 in
length = 16 - 2(2) = 12 in
width = 10 - 2(2) = 6