<p>The perimeter of an isosceles triangle of base length $x$ cm and two sides of equal length $y$ cm is $10$ cm.</p>
<p>By using differentiation, show that when the area of the triangle is maximum,</p>
<ol class="abc-list"><li><div class="list-item-paragraph">the triangle is equilateral,</div></li><li><div class="list-item-paragraph">the area is $\frac{25\sqrt{3}}{9}$ units$^{2}$.</div></li></ol>

The perimeter of an isosceles triangle of base length $x$ cm and two sides of equal length $y$ cm is $10$ cm.

By using differentiation, show that when the area of the triangle is maximum,

the area is $\frac{25\sqrt{3}}{9}$ units$^{2}$.

Default solution

Two variables: $x$ & $y$, fixed condition: 10 cm perimeter

\[x + 2y = 10 \Rightarrow y = \dfrac{10 - x}{2}\]

\[h^2 + \left(\dfrac{x}{2}\right)^2 = y^2 \Rightarrow h^2 = y^2 - \dfrac{1}{4}x^2 \Rightarrow h = \sqrt{y^2 - \dfrac{1}{4}x^2}\]

Area, $A = \dfrac{1}{2} \times x \times h = \dfrac{1}{2}x\sqrt{y^2 - \dfrac{1}{4}x^2}$ [2 variables]

\[A = \dfrac{1}{2}x\sqrt{\left(\dfrac{10-x}{2}\right)^2 - \dfrac{1}{4}x^2}\]

\[A = \dfrac{1}{2}x\sqrt{\dfrac{(10-x)^2}{4} - x^2}\]

\[A = \dfrac{1}{2}x \cdot \dfrac{1}{2}\sqrt{(10-x)^2 - x^2}\]

\[A = \dfrac{1}{4}x\sqrt{(10-x)^2 - x^2}\] [1 variable]

\[A = \dfrac{1}{4}x\sqrt{(10-x+x)(10-x-x)} = \dfrac{1}{4}x\sqrt{10(10-2x)}\]

\[A = \dfrac{1}{4}x\sqrt{10 \cdot 2(5-x)} = \dfrac{1}{4}x\sqrt{4 \cdot 5}\sqrt{5-x}\]

\[A^2 = \dfrac{5}{4}x^2(5-x) = \dfrac{5}{4}(5x^2 - x^3)\]

\[2A\dfrac{\mathrm{d}A}{\mathrm{d}x} = \dfrac{5}{4}(10x - 3x^2)\]

At stationary value, $\dfrac{\mathrm{d}A}{\mathrm{d}x} = 0$

\[\dfrac{5}{4}(10x - 3x^2) = 0 \Rightarrow x(10 - 3x) = 0\]

\[x = \dfrac{10}{3} \text{ or } x = 0 \text{ (rej as } x > 0\text{)}\]

\[2A\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2} + 2\left(\dfrac{\mathrm{d}A}{\mathrm{d}x}\right)^2 = \dfrac{5}{4}(10 - 6x)\]

From 0, at stationary point: $2A\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2} + 0 = \dfrac{5}{4}\left(10 - 6 \cdot \dfrac{10}{3}\right)$

\[\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2} = \dfrac{-25}{4A} < 0 \text{ since } A > 0\]

\[x = \dfrac{10}{3}, \quad y = \dfrac{10-x}{2} = \dfrac{10}{3} \text{ cm}\]

Since $x = y$, thus the triangle is equilateral.

\[A = \dfrac{\sqrt{5}}{2}\left(\dfrac{10}{3}\right)\sqrt{5 - \dfrac{10}{3}} = \dfrac{5}{3}\sqrt{5} \cdot \sqrt{\dfrac{5}{3}}\]

\[= \dfrac{5}{3}\sqrt{5} \cdot \dfrac{\sqrt{5}}{\sqrt{3}} = \dfrac{25}{3\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{25\sqrt{3}}{9} \text{ units}^2\]

Worked Example 6

Worked Example 6

Video Solution:

Default solution