Starting from the initial trial solution (x₁,x₂)= (0,0),interactively apply the gradient search procedure with $\varepsilon$ = 1 to solve (approximately)the following problem.Minimize g(x)= 2 + - - 2x₁x₂.

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The Answer is For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 $For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 - + 4x1 + 2x1x2.Taking the partial derivatives,we have , .Since \varepsilon = 1,we will stop the gradient search procedure when the current trial solution yields ≤ 1 for j = 1 and 2.Starting from (x1,x2)= (0,0),the gradient search procedure proceeds as summarized in the following table: At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x1 = 1.5 and x2 = 1 as the desired approximation of an optimal solution (which happens to be x1 = 2 and x2 = 2).To elaborate further on how the above table was obtained,at iteration 1, = f (4t,0)= -2(4t)2 - 02 + 4(4t)+ 2(4t)(0)= -32t2 + 16t.Because f (4t*,0)= and = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.$ - $For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 - + 4x1 + 2x1x2.Taking the partial derivatives,we have , .Since \varepsilon = 1,we will stop the gradient search procedure when the current trial solution yields ≤ 1 for j = 1 and 2.Starting from (x1,x2)= (0,0),the gradient search procedure proceeds as summarized in the following table: At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x1 = 1.5 and x2 = 1 as the desired approximation of an optimal solution (which happens to be x1 = 2 and x2 = 2).To elaborate further on how the above table was obtained,at iteration 1, = f (4t,0)= -2(4t)2 - 02 + 4(4t)+ 2(4t)(0)= -32t2 + 16t.Because f (4t*,0)= and = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.$ + 4x₁ + 2x₁x₂.Taking the partial derivatives,we have $For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 - + 4x1 + 2x1x2.Taking the partial derivatives,we have , .Since \varepsilon = 1,we will stop the gradient search procedure when the current trial solution yields ≤ 1 for j = 1 and 2.Starting from (x1,x2)= (0,0),the gradient search procedure proceeds as summarized in the following table: At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x1 = 1.5 and x2 = 1 as the desired approximation of an optimal solution (which happens to be x1 = 2 and x2 = 2).To elaborate further on how the above table was obtained,at iteration 1, = f (4t,0)= -2(4t)2 - 02 + 4(4t)+ 2(4t)(0)= -32t2 + 16t.Because f (4t*,0)= and = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.$ , $For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 - + 4x1 + 2x1x2.Taking the partial derivatives,we have , .Since \varepsilon = 1,we will stop the gradient search procedure when the current trial solution yields ≤ 1 for j = 1 and 2.Starting from (x1,x2)= (0,0),the gradient search procedure proceeds as summarized in the following table: At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x1 = 1.5 and x2 = 1 as the desired approximation of an optimal solution (which happens to be x1 = 2 and x2 = 2).To elaborate further on how the above table was obtained,at iteration 1, = f (4t,0)= -2(4t)2 - 02 + 4(4t)+ 2(4t)(0)= -32t2 + 16t.Because f (4t*,0)= and = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.$ .Since

\varepsilon

= 1,we will stop the gradient search procedure when the current trial solution yields $For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 - + 4x1 + 2x1x2.Taking the partial derivatives,we have , .Since \varepsilon = 1,we will stop the gradient search procedure when the current trial solution yields ≤ 1 for j = 1 and 2.Starting from (x1,x2)= (0,0),the gradient search procedure proceeds as summarized in the following table: At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x1 = 1.5 and x2 = 1 as the desired approximation of an optimal solution (which happens to be x1 = 2 and x2 = 2).To elaborate further on how the above table was obtained,at iteration 1, = f (4t,0)= -2(4t)2 - 02 + 4(4t)+ 2(4t)(0)= -32t2 + 16t.Because f (4t*,0)= and = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.$ ≤ 1 for j = 1 and 2.Starting from (x₁,x₂)= (0,0),the gradient search procedure proceeds as summarized in the following table: $For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 - + 4x1 + 2x1x2.Taking the partial derivatives,we have , .Since \varepsilon = 1,we will stop the gradient search procedure when the current trial solution yields ≤ 1 for j = 1 and 2.Starting from (x1,x2)= (0,0),the gradient search procedure proceeds as summarized in the following table: At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x1 = 1.5 and x2 = 1 as the desired approximation of an optimal solution (which happens to be x1 = 2 and x2 = 2).To elaborate further on how the above table was obtained,at iteration 1, = f (4t,0)= -2(4t)2 - 02 + 4(4t)+ 2(4t)(0)= -32t2 + 16t.Because f (4t*,0)= and = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.$ At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x₁ = 1.5 and x₂ = 1 as the desired approximation of an optimal solution (which happens to be x₁ = 2 and x₂ = 2).To elaborate further on how the above table was obtained,at iteration 1, $For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 - + 4x1 + 2x1x2.Taking the partial derivatives,we have , .Since \varepsilon = 1,we will stop the gradient search procedure when the current trial solution yields ≤ 1 for j = 1 and 2.Starting from (x1,x2)= (0,0),the gradient search procedure proceeds as summarized in the following table: At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x1 = 1.5 and x2 = 1 as the desired approximation of an optimal solution (which happens to be x1 = 2 and x2 = 2).To elaborate further on how the above table was obtained,at iteration 1, = f (4t,0)= -2(4t)2 - 02 + 4(4t)+ 2(4t)(0)= -32t2 + 16t.Because f (4t*,0)= and = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.$ = f (4t,0)= -2(4t)² - 0² + 4(4t)+ 2(4t)(0)= -32t² + 16t.Because f (4t*,0)= $For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 - + 4x1 + 2x1x2.Taking the partial derivatives,we have , .Since \varepsilon = 1,we will stop the gradient search procedure when the current trial solution yields ≤ 1 for j = 1 and 2.Starting from (x1,x2)= (0,0),the gradient search procedure proceeds as summarized in the following table: At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x1 = 1.5 and x2 = 1 as the desired approximation of an optimal solution (which happens to be x1 = 2 and x2 = 2).To elaborate further on how the above table was obtained,at iteration 1, = f (4t,0)= -2(4t)2 - 02 + 4(4t)+ 2(4t)(0)= -32t2 + 16t.Because f (4t*,0)= and = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.$ and $For convenience,we can convert from minimization to maximization by considering f(x)= -g(x).(Section 13.5 in the textbook also describes how the procedure can be adjusted to perform minimization directly.)Thus,the equivalent problem becomes Maximize f(x)= -2 - + 4x1 + 2x1x2.Taking the partial derivatives,we have , .Since \varepsilon = 1,we will stop the gradient search procedure when the current trial solution yields ≤ 1 for j = 1 and 2.Starting from (x1,x2)= (0,0),the gradient search procedure proceeds as summarized in the following table: At the beginning of iteration 4,both partial derivatives (0 and 1)are ≤ 1,so the procedure stops with x1 = 1.5 and x2 = 1 as the desired approximation of an optimal solution (which happens to be x1 = 2 and x2 = 2).To elaborate further on how the above table was obtained,at iteration 1, = f (4t,0)= -2(4t)2 - 02 + 4(4t)+ 2(4t)(0)= -32t2 + 16t.Because f (4t*,0)= and = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.$ = -64t + 16 = 0,it follows that t* = 0.25.The calculations for iterations 2 and 3 follow the same pattern.

Starting from the Initial Trial Solution (X1,x2)= (0,0),interactively Apply the Gradient

Starting from the Initial Trial Solution (X₁,x₂)= (0,0),interactively Apply the Gradient