The lateral edge of a regular hexagonal pyramid is 1 cm. If the volume is maximum, then its height must be equal to :

Question

The lateral edge of a regular hexagonal pyramid is 1 cm. If the volume is maximum, then its height must be equal to : (A) (1/3) (B) (2/3) (C) (1/√3) (D) 1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/93ba5dc2d4146afc596eeb6a7b77f2a2-.png" /> x 2+ h 2=1 ; V =(1/3) ⋅ 6 ⋅ (√3/4) x 2 h =(√3/2) h (1- h 2) V prime( h )=0 ⇒ h =(1/√3) ⇒ V max =1 / 3

Q. The lateral edge of a regular hexagonal pyramid is $1 c m$ . If the volume is maximum, then its height must be equal to :

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{1}{3}$

1

$x^{2} + h^{2} = 1; V = \frac{1}{3} \cdot 6 \cdot \frac{3}{4} x^{2} h = \frac{3}{2} h (1 - h^{2})$
$V^{'} (h) = 0 \Rightarrow h = \frac{1}{3} \Rightarrow V_{m a x} = 1/3$

Q. The lateral edge of a regular hexagonal pyramid is 1cm. If the volume is maximum, then its height must be equal to :

31​

32​

3​1​

1

x2+h2=1;V=31​⋅6⋅43​​x2h=23​​h(1−h2) V′(h)=0⇒h=3​1​⇒Vmax​=1/3

Q. The lateral edge of a regular hexagonal pyramid is $1 c m$ . If the volume is maximum, then its height must be equal to :

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{1}{3}$

$x^{2} + h^{2} = 1; V = \frac{1}{3} \cdot 6 \cdot \frac{3}{4} x^{2} h = \frac{3}{2} h (1 - h^{2})$
$V^{'} (h) = 0 \Rightarrow h = \frac{1}{3} \Rightarrow V_{m a x} = 1/3$