Question

. If you have multiple pages, please staple them together. 1. (a) Eind the value of r that maximizes the following function. Note that z20. Hint: For ax2 + bx + c = 0, the values of x which are the solutions of the equation are given by: 2a (b) At the value of you found in (a), does it satisfy the second-order condition for the maximum? Explain your answer.

Answer 1

a) To find the maximum value we have equate

f'(x) = 0

f(x) = 10x - ( 4x + 5 -2x^{​​​​2}​​​ + (1/3)x^{​​​​​​3}​)

So differentiate f(X) wrt X

f'(x) = 10 - (4-4x + 2x²) = -x2 + 4x +6

So now

f'(x) = 0

-x^{​​​​​​2} +4x +6 = 0

x^{​​​​​2}​​​​​ -4x -6 = 0

x =( -(-4) $\pm$ sqrt (4² - 4(1*-6) )/ (2*1)

x = (4$\pm$6.32)/2

x = 2.064 or -1.16

Maximum value of X = 2.064

2) Second order condition of maximum

f"(X) = 0

f'(X) = -x^{​​​​​2}​​​​+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)