Question

Please answer the questions using MATLAB

Exercise 1 Dimensions of the Largest Box An open bols to be made rom ฮ rectangular poce of cardboard measuring 8 x48. The box s made by cutting o ual squares rom cach of its 4 corners and turning up the sides. Suggestion: you can try making one yourself with of paper) spare piece u8. 1. Let x be the side of a square removed from each corner. Express the volume v of the box as a function of x. (The length is 18-2x, what is the width?) Enter a short answer 2. What is the domain of values for x? the domain of Wx Enter a short answer nte your answer in interval notation. ea. (a.bl instead of a xb 3. Make a graph of ix)over the domain chosen. Graphically determine all possible values of x so that the volume of the hox is V-1400 cubic inches. Determine these values to an acruracy of 3 signiticant digits Enter a list of numbers separated by space or commas. uation V ㄨ-1400 is cubic equa tion with 3 real roots. Find all three roots using MATLAB's "roats. corrimand. Du you cet the samle answers as your gra phi invesbgat on. Exp air 5. Graph Vx) and)tagcther, lsc the graph of V'(x) to find the valuc of x that maximizes V Enter a number l a show that your last answer is a critical point for Vx)? 6. Find V(x) analytically by differentiating your formula for Vx). Does your formu ect exactly one of the choices. yes no 7. Find the exact value of the maximum of Vx Enter a number

Answer 1

Please find attached the screenshots of the Matlab graphs and script at the end. Run the script to verify the Matlab generated answers.

The length is 48-2x , width is 18-2x, and height is herepore, the volume V oF the box as o functon of x is given as, v(x) = (48-2x)(18-2x) χ 4χ3-1322+86 4 χ The domain OF Possible values of χ IS 2. (o, 9] The screenshot of the graph is ottached below (at the end) The values of x estimated graphicaly so that volume of boz is 1400 cubic inches are 2.504 inch and 5.594 Inch respectively. 4.Vx) -1400 Rewrite the equation as V(x)-1 400 0 From, part (1) V(X)s 4χ3-132x +864 χ Substitute in the equation 0 4x. 132 χ + 864-1400 Therefore, P- L4, -132, 864, -1400 = Toots (p) hus, roots ore given as 24.8854, s.6056, 2.5090 since our graph domain is limited from [o, 9J,the vefoxe, root 24,8854 was undetermined thugh gro hicol nvestoh

5. The plots of V(x) vs χ and V,1%) vs x are attached at the end. Through graph the estimated value of x that haximises V is 3.992 v'(x): 12z?-264 χ +864 o determine the critical paint, Find root of the equahion (X2-222+72):0 12 (x2 -18x- 42 +72) 0 12 [2(x-18)-4 (x-18)] כ 12 0 12 (x-4)(x-18) x 4, 18 neglect χ:18 Since, xe (0,9) Hence our Formula shows last answer is a critical paint For V'(x), ie., answer is yes Vx), ie, answer is yes. 7. From part (6) at x4 VCz) is maximum,so, substitule X s4 in the expessian for Vx) V(r 4) 4X43- 132 X41+864x4 1600 hus, the exact value oximum of Viz) is 1600 cubic inches

part (3)

1600 1400 X: 2.504 Y: 1399 1200 1000 800 600 400 200 3 4 5 6 7 9

1600 1400 X: 5.594 Y: 1403 1200 1000 800 600 400 200 3 4 5 6 7 9

part (5)

2000 dV/dx 1500 X: 3.992 Y: 1600 1000 500 -500 -1000 3 4 5 6 7 9

Matlab Script

clear; close all; clc;

%% part 3

x = linspace( 0, 9, 400 );
V = (48-2*x).*(18-2*x).*x
%V = 4*x.^3 - 132*x.^2 + 864*x ;

figure
plot( x, V )
xlabel( '\bf x' )
ylabel( '\bf V(x)' )
grid on

%% part 4

syms symX

Volume = expand( (48-2*symX) * (18-2*symX) * symX )
VolumePrime = simplify( diff( Volume ) )

p = [ 4 -132 864 -1400 ];
r = roots(p)

%% part 5

Vprime = 12*x.^2 - 264*x + 864 ;

figure
plot( x, V, x, Vprime )
xlabel( 'x' )
legend( 'V(x)', 'dV/dx' )
grid on