Question

A fence is required around a field is shaped as shown below. It consists of a rectangle of length Land width Wand a right triangle that is symmetrical about the central horizontal axis of the rectangle. Suppose the width is known (in metres), and the enclosed area A is known (in square metres). w 1. Use pen and paper to determine the equations for the total area and perimeter in terms of the width W, and length L. 2. Use MATLAB to plot the perimeter against the width as a black solid line, assuming the width to be between 7 to 20 metres and the area to be 111 m². 3. Use the min() function to determine the minimum perimeter required to fence off the 111 m2 area. Print the corresponding length and width required, and mark the minimum point on the previous plot with a blue diamond. Note: Try solving/rearranging the equations by hand before putting them into MATLAB Note: You may want to use the built-in function 'min' with two outputs

Answer 1

============ (1) ============

Let us calculate the area and periphery first.

As the right angle trianlge is symmetrical about the central horizontal axis, observe the following:

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It is to be noted that, the value of the triangle dimension is calculated using pythagoras theorem (as the triangle is a right angled triangle)

This gives the total periphery as:

$Perimeter\,(P) = W+2L + 2\left ( \frac{W}{\sqrt{2}} \right )=W(\sqrt{2}+1)+2L$

$Area\,(A) = (W\times L)+\left [ \frac{1}{2}\times \frac{W}{\sqrt{2}}\times \frac{W}{\sqrt{2}} \right ]=\frac{W^2}{4}+WL$

These formulations will be used int he next subsections:

============ (2) and (3) ============

Please find the required MATLAB script for the two pars as the following:

%==========================================================================

clear all;clc;

%================================= (2) =================================
% Given data:
W = 7:0.1:20; % Given width range
A = 111;

% Calculating Length (L) as from the calculated Area formulation:
L = (A - ((W.^2)/4))./W;

% Calculating perimeter as:
P = W*(sqrt(2) + 1) + (2*L);

figure
plot(W,P,'k','LineWidth',1.2)
hold on
xlabel('Width (W) meters')
ylabel('Perimeters (P) meters')
title('Width (W) v/s Perimeter (P)');
xlim([min(W) max(W)])
grid on

%================================= (3) =================================
% Finding minmum paremeter

min_Perimeter_Index = find(P==min(P));
min_Perimeter = P(min_Perimeter_Index);
min_Peri_Length = L(min_Perimeter_Index);
min_Peri_Width = W(min_Perimeter_Index);
plot(min_Peri_Width,min_Perimeter,'bd','MarkerSize',10);
fprintf('To fence off the 111 m^2 are, the minimum perimeter (P) = %f\n',min_Perimeter);
fprintf('Corresponding Length (L)= %f and Width (W) = %f\n',min_Peri_Length,min_Peri_Width);

%==========================================================================

output:

To fence off the 111 m^2 are, the minimum perimeter (P) = 41.229062
Corresponding Length (L)= 7.577778 and Width (W) = 10.800000

Width (W) vls Perimeter (P) Perimeters (P) meters 7 8 9 10 11 16 17 18 19 20 12 13 14 15 Width (W) meters

Note: Please find the diamond marker at the bottom indicating minimum Width value corresponding to minimum Perimeter.

Hope this helps!

In case of any clarification, please comment.

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