As per HOMEWORKLIB RULES first question is answered
Answer:
a) To find the maximum value we have equate
f'(x) = 0
f(x) = 10x - ( 4x + 5 -2x2 + (1/3)x3)
So differentiate f(X) wrt X
f'(x) = 10 - (4-4x + 2x2) = -x2 + 4x +6
So now
f'(x) = 0
-x2 +4x +6 = 0
x2 -4x -6 = 0
x =( -(-4) sqrt (42 - 4(1*-6) )/ (2*1)
x = (46.32)/2
x = 2.064 or -1.16
Maximum value of X = 2.064
2) Second order condition of maximum
f"(X) = 0
f'(X) = -x2+4x+6
f"(X) = -2x + 4
f"(X) = 0
-2x +4 =0
x = 2
Yes the value for second order is satisfying the first order
maximum value (approximately)
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