By Dr Verda Salman

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Optimization- Minimum/Maximum ValuesGoal equilibrium“quest for the best”

Constant function

Monotonically increasing/decreasing function

Implications of second derivatives

Functions with relative maxima and minima

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Minimum/Maximum Values- First Derivative Test

• The relative maximum: if a derivative f’(x) changes its sign from positive to negative from the immediate left of the point x0 to its immediate right.

• A relative Minimum: if a derivative f’(x) changes its sign from negative to positive from the immediate left of the point x0 to its immediate right.

• Stationary Points: Points when f’(x)=0

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Necessary and Sufficient Condition for First Derivative Test of Optimization

Necessary Condition: f’(X) = 0 is satisfied

Sufficient Condition: Change of derivative sign before and after extrema.

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Relative extrema of a functionFind the relative extrema of the following

function (X>0)

Y = f(x) = x3 -12x2 +36x +8dy/dx = 3X2 – 24x + 36Putting dy/dx = 0X = 2, 6F(2) = 40 relative maximumF(6) = 8 relative minimum

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Second Derivative Test

A relative maximum if the second-derivative value at x0 is f’’(x0)<0

A relative minimum if the second-derivative value at x0 is f’’(x0)>0

Necessary and Sufficient Condition for Second Derivative Test of Optimization

Necessary Condition: f’(X) = 0 is satisfied

Sufficient Condition: Second derivative must have a sign to confirm concavity or convexity.

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Second Derivative Test

Examples:Find stationary values maximum/minimum

points with second derivative testY = 4X2 – XY = X3 – 3X2 + 2

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Constrained Optimization: Lagrange-Multiplier Method (choosing the best solution)Maximize: U = x1x2 + 2x1

Subject to : 4x1 + 2X2 = 60Step 1: Write a new function ZZ = x1x2 + 2x1 + ג( 60-4 x1 - 2X2 )

Step 2: Derivative of Z w.r.t ג, x1, x2

Z ג = бZ/б 60-4 = ג x1 - 2X2

Zx1 = бZ/бX1 = X2 + 2 - 4ג

ZX2 = бZ/бX2 = X1 - 2ג 8