Answer to Question #291521 in Calculus for Kanh

Question

The perimeter of a triangle is 60cm. Find the length of the sides of the triangle which gives a maximum area.

Answer 1

Let x, y, z be the sides of a triangle.

Then, the perimeter of the triangle is;

$\displaystyle p(x,y,z)=x+y+z$

Also, using Heron's formula, the area of the triangle is;

$\displaystyle A(x,y,z)=\frac{\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}}{4}$

Now, from the question, $\displaystyle P(x,y,z)=60$. Thus, we are to;

maximize: $\displaystyle A(x,y,z)=\frac{\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}}{4}$

subject to: $\displaystyle x+y+z-60=0$

Using Lagrange's method;

$\displaystyle \Rightarrow\nabla\left(\frac{\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}}{4}\right)=\lambda\nabla(x+y+z-60)\\ \Rightarrow\quad\frac{4xy^2+4xz^2-4x^3}{8\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}}=\lambda\\ \qquad \frac{4yx^2+4yz^2-4y^3}{8\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}}=\lambda\\ \qquad \frac{4zx^2+4zy^2-4z^3}{8\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}}=\lambda\\ \quad\\ \text{Thus, we need to solve the following equations:}\\ \quad\\ 4xy^2+4xz^2-4x^3=\lambda8\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}\ \cdots\cdots\cdots\cdots\cdots\cdots(i)\\ 4yx^2+4yz^2-4y^3=\lambda8\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}\ \cdots\cdots\cdots\cdots\cdots\cdots(ii)\\ 4zx^2+4zy^2-4z^3=\lambda8\sqrt{2x^2y^2+2x^2z^2+2y^2z^2-x^4-y^4-z^4}\ \cdots\cdots\cdots\cdots\cdots\cdots(iii)\\ x+y+z-60=0\cdots\cdots\cdots\cdots\cdots\cdots(iv)$

substituting (i) into (ii) and (iii) yields;

$\displaystyle 4yx^2+4yz^2-4y^3=4xy^2+4xz^2-4x^3\\ 4zx^2+4zy^2-4z^3=4xy^2+4xz^2-4x^3\\ x+y+z-60=0\\ \Rightarrow\\ yx^2+yz^2-y^3-xy^2-xz^2+x^3=0\\ zx^2+zy^2-z^3-xy^2-xz^2+x^3=0\\ x+y+z-60=0\\ \Rightarrow\\ xy(x-y)-z^2(x-y)+(x^3-y^3)=0\\ xz(x-z)-y^2(x-z)+(x^3-z^3)=0\\ x+y+z-60=0\\ \quad\\ \text{But }(a^3-b^3)=(a-b)(a^2+ab+b^2)\\ \Rightarrow\\ xy(x-y)-z^2(x-y)+(x-y)(x^2+xy+y^2)=0\\ xz(x-z)-y^2(x-z)+(x-z)(x^2+xz+z^2)=0\\ x+y+z-60=0\\ \Rightarrow\\ (x-y)[xy-z^2+(x^2+xy+y^2)]=0\\ (x-z)[xz-y^2+(x^2+xz+z^2)]=0\\ x+y+z-60=0\\ \Rightarrow\\ (x-y)[x^2+x(2y)+(y^2-z^2)]=0\\ (x-z)[x^2+x(2z)+(z^2-y^2)]=0\\ x+y+z-60=0\\ \Rightarrow\\ (x-y)(x+y-z)(x+y+z)=0\\ (x-z)(x+z-y)(x+z+y)=0\\ x+y+z-60=0\\ \Rightarrow\\ (y-z)\times z\times(2y+x)=0\\ 2y+z-60=0,\ \text{if }x=y\\ \Rightarrow\\ 3z-60=0,\ \text{if }y=z\\ \Rightarrow\\ x=y=z=20$

Hence, the length of the sides of a triangle of perimeter 60cm which gives a maximum area are;

$\displaystyle x=y=z=20\ cm$, where x, y, z are the sides of the triangle.