The sum of the base and altitude of a triangle is 30 cm. What is the maximum possible area of such a triangle?
100 cm$^2$
110 cm$^2$
112.5 cm$^2$
120 cm$^2$
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CAPF – 2013
The area of a triangle is given by $A = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2}bh$.
From the given condition, $h = 30 – b$. Substitute this into the area formula:
$A = \frac{1}{2} b (30 – b) = \frac{1}{2} (30b – b^2) = 15b – \frac{1}{2}b^2$.
This is a quadratic function of $b$ in the form $Ab^2 + Bb + C$, where $A = -\frac{1}{2}$, $B = 15$, and $C = 0$. Since the coefficient of $b^2$ is negative, the parabola opens downwards, and its maximum value occurs at the vertex.
The vertex of a parabola $ax^2 + bx + c$ is at $x = -\frac{B}{2A}$. In this case, $b = -\frac{15}{2(-\frac{1}{2})} = -\frac{15}{-1} = 15$.
The maximum area occurs when the base $b = 15$ cm.
When $b = 15$, the altitude $h = 30 – b = 30 – 15 = 15$ cm.
The maximum possible area is $A = \frac{1}{2} \times 15 \times 15 = \frac{1}{2} \times 225 = 112.5$ cm$^2$.
Given $b+h=30$, we have $\frac{30}{2} \ge \sqrt{bh} \implies 15 \ge \sqrt{bh}$. Squaring both sides gives $225 \ge bh$. The maximum value of $bh$ is 225.
The area $A = \frac{1}{2}bh$, so the maximum area is $\frac{1}{2} \times 225 = 112.5$. Equality in AM-GM holds when $b=h$, confirming the previous result.