ps6solnsW06

# ps6solnsW06 - %Engineering 6 Winter 2006 Problem 6.1...

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%Engineering 6, Winter 2006, Problem 6.1 Solution %(Calculating max volume of tray without calculus) %Suppress extra lines in output and set fixed short display format compact ; format short ; clc; %Clear command window and variables clear; %The volume of the tray is: V = (14.5-2x)(6-2x)(x) %(for a length of 14.5, width of 6) %The values of x can range from 0 up to 6/2 = 3 m (if x is larger than 3, %the foldup sides disappear on the 6 meter side of the metal piece). %Taking values of x at intervals of 0.02 means that our answer %will be accurate to 0.02 meters. %Define dimensions of metal plate length = 14.5; width = 6; %Define vector of x values and calculate corresponding volumes x = 0:0.02:width/2; V = (length-2*x).*(width-2*x).*x; %Use max command to find the max volume and the index (k_max) where it occurs [V_max, k_max] = max(V); %Find the value of x that corresponds to V_max x_max = x(k_max); %The tray height will be x_max. Calculate tray length and width. traylength = length - 2*x_max;

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## This note was uploaded on 11/02/2009 for the course EEC eng 6 taught by Professor Lagerstrom during the Winter '06 term at UC Davis.

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ps6solnsW06 - %Engineering 6 Winter 2006 Problem 6.1...

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