Answer:
Step-by-step explanation:
Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.
The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.
Substituting 114 - 2s for h in the volume formula, we obtain:
V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)
This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:
dV
----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2
ds
Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.
Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in
and the volume is V = s^2(h) = 46,298.25 in^3
Answer:
square of longest side ≠ the sum of the squares of the other two sides
It's not a right triangle
Step-by-step explanation:
The converse of the Pythagorean Theorem is: If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
square of longest side: 12 12² = 144
the sum of the squares of the other two sides: 8² + 10² = 164
Answer:
in steps
Step-by-step explanation:
DE // BC
m∠ADE = m∠ABC and m∠AED = m∠ACB
∴ ΔADE similar to ΔABC
AB/AD = AC/AE
(AD + DB) / AD = (AE + EC) / AE
AD/AD + DB/AD = AE/AE + EC/AE
1 + DB/AD = 1 + EC/AE
DB/AD = EC/AE (AD/DB = AE/EC)
