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# Find the height of a pyramid when its volume maximum

Question:

P-ABCD is a regular square pyramid. PA = 6. When the volume is the largest, what is the height? Solution:

The volume of a pyramid is: V = (1/3) B h, in which B is the base area and h is the height of the pyramid.

Because P-ABCD is a regular square pyramid, so the base ABCD is a square, so, the base area of the pyramid is: B = AB × BC = AB2, h is the height of the pyramid. Connect AC and BD, AC and BD are two diagonals, and AC and BD intersect at the point O. Because P-ABCD is a regular square pyramid, so, the point O is the projection of the point P on the plane ABCD. Because PO is perpendicular to plane ABCD and AO lies in the plane ABCD, so, PO is perpendicular to AO. So, the angle POA is a right angle, so, the triangle POA is a right triangle.

Note: h is the height of the pyramid. Let h = PO

In right triangle POA,
AO2 = PA2 – PO2
= 36 – h2

Now, we find the relation between AO and AB Because ABCD is a square, AC and BD are two diagonals, so, AC = BD and AC and BD bisect each other and perpendicular to each other at point O. So, angle AOB = 90 degrees, AO = OB, angle OAB = angle OBA = 45 degrees.

In right triangle OAB,
AB2 = OA2 + OB2
AB2 = 2 OA2
AB = (square root of 2) × OA.
The base area of the pyramid is:
B = AB2 = 2 (36 – h2)
the volume of the pyramid is:
V = (1/3) B h
= (1/3) x 2 x (36 – h2) x h
= (2/3) (-h3 + 36h)

The volume V is the function of the height of the pyramid

V(h) = (2/3) (-h3 + 36h)

In right triangle POA, h = PO is a leg, PA is the hypotenuse, so h is in the range of (0 < h < 6)

Find the first derivative of the function V(h)

dv/dh = (2/3) (-3h2 + 36)
= 2 (-h2 + 12)

When the first derivative of the function V(h) is greater than zero, the function V(h) increase.

dv/dh > 0
-h2 + 12 > 0
h2 < 12

When 0 < h < 2 × (square root of 3), the function v(h) increase

When h = 2 × (square root of 3), the function v(h) has the maximum value

When 2 × (square roots of 3) < h < 6, function v(h) decrease.