Maximum area of parabolic section
$A_{max} = \frac{2}{3}(2b)h$
$207.8 = \frac{4}{3}bh$
$h = \dfrac{155.85}{b}$ → Equation (1)
Triangle DEC is similar to triangle ABC
$\dfrac{h}{24 - x} = \dfrac{L}{24}$
$h = \frac{1}{24}L(24 - x)$ → Equation (2)
For the right triangle at the base of the cone
$b^2 + (12 - x)^2 = 12^2$
$b^2 = 12^2 - (12 - x)^2$
$b^2 = 12^2 - (12^2 - 24x + x^2)$
$b = \sqrt{24x - x^2}$ → Equation (3)
Area of parabolic section at any distance x
$A = \frac{2}{3}(2b)h$
$A = \frac{4}{3}\sqrt{24x - x^2}[ \, \frac{1}{24}L(24 - x) \, ]$
$A = \frac{1}{18}L(24 - x)\sqrt{24x - x^2}$
$\dfrac{\partial A}{\partial x} = \frac{1}{18}L \left[ (24 - x) \dfrac{24 - 2x}{2\sqrt{24x - x^2}} + \sqrt{24x - x^2} ~ (-1) \right] = 0$
$\dfrac{2(12 - x)(24 - x)}{2\sqrt{24x - x^2}} - \sqrt{24x - x^2} = 0$
$\dfrac{(12 - x)(24 - x)}{\sqrt{24x - x^2}} = \sqrt{24x - x^2}$
$(12 - x)(24 - x) = 24x - x^2$
$288 - 36x + x^2 = 24x - x^2$
$2x^2 - 60x + 288 = 0$
$x^2 - 30x + 144 = 0$
$x = 24 \, \text{ and } \, 6$
Use
$x = 6 \, \text{ cm}$
Thus,
$b = \sqrt{24x - x^2}$ → From Equation (3)
$b = \sqrt{24(6) - 6^2}$
$b = 6\sqrt{3} \, \text{ cm}$
Base-width of parabolic section
$2b = 12\sqrt{3}$
$2b = 20.78 \, \text{ cm}$ Part 1: [ D ]
Altitude of parabolic section
$h = \dfrac{155.85}{b}$ → From Equation (1)
$h = \dfrac{155.85}{6\sqrt{3}}$
$h = 15 \, \text{ cm}$ Part 2: [ C ]
From
$h = \frac{1}{24}L(24 - x)$ → From Equation (2)
$15 = \frac{1}{24}L(24 - 6)$
$L = 20 \, \text{ cm}$
By Pythagorean theorem
$H^2 + 12^2 = L^2$
$H^2 + 12^2 = 20^2$
$H = 16 \, \text{ cm}$ Part 3: [ C ]
Note:
For maximum area of parabola that can be cut from a right circular cone:
$b = D \sin 60^\circ$
$h = 0.75L$
$x = 0.25D$
Solving the problem using the above formulas