Thus the only critical point inside d is 0 0 and at

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Thus, the only critical point inside D is (0 , 0) and at this critical point f (0 , 0) = 3 . On the other hand, the boundary of D con- sists of the line segments L 1 , L 2 , L 3 , and L 4 , as shown above. (i) L 1 = { ( x, 1) : 1 x 1 } , (ii) L 2 = { (1 , y ) : 1 y 1 } , (iii) L 3 = { ( x, 1) : 1 x 1 } , (iv) L 4 = { ( 1 , y ) : 1 y 1 } . But on L 1 , f ( x, y ) = 6 x 2 + 8 2 x 2 + 3 = 4 x 2 + 11 while on L 2 , f ( x, y ) = 6 + 8 y 2 + 2 y + 3 = 8 y 2 + 2 y + 9 , on L 3 , f ( x, y ) = 6 x 2 + 8 + 2 x 2 + 3 = 8 x 2 + 11 and on L 4 , f ( x, y ) = 6 + 8 y 2 + 2 y + 3 = 8 y 2 + 2 y + 9 . Thus (i) on L 1 abs max value of f is 15 , (ii) on L 2 abs max value of f is 19 , (iii)on L 3 abs max value of f is 19 , (iv)on L 4 abs max value of f is 19 . Consequently, taking the largest of 15 , 19 , 19 , 19 , 3 , we see that on D abs max value of f is 19 . 043 10.0points A rectangular box x y
sikdar (rs52326) – Practice for midterm 2 – perutz – (54175) 24 with no top is to be constructed having a volume of 32 cubic inches. What length y , in inches, will minimize the amount of material to be used in its construction. (Assume no material is wasted.) 1. length = 3 inches 2. length = 5 inches 3. length = 4 inches correct 4. length = 7 inches 5. length = 6 inches Explanation: When the box has width x , length y and height z , the material A in the box is given by A ( x, y, z ) = xy + 2 xz + 2 yz since the box has no top. On the other hand, the box has volume 32 cubic inches, so xyz = 32 . Eliminating z from these two equations we thus see that A ( x, y ) = xy + 64 y + 64 x . The minimum value of A occurs at a critical point. Now, after differentiation, A x = y 64 x 2 , A y = x 64 y 2 . The critical point of A thus occurs at the solution of the equations y = 64 x 2 , x = 64 y 2 , i.e. , when x = y = 4. To check this gives a local minimum of A ( x, y ), observe that A = A xx (4 , 4) = 128 x 3 vextendsingle vextendsingle vextendsingle x =4 = 2 > 0 , C = A yy (4 , 4) = 128 y 3 vextendsingle vextendsingle vextendsingle y =4 = 2 > 0 , and B = A xy (4 , 4) = 1 , so AC B 2 > 0 at the critical point. Conse- quently, A = A ( x, y ) has a local minimum at (4 , 4), showing that the least amount of ma- terials will be used in the construction of the box when it has length = 4 inches . 044 10.0points Locate the point at which the function f ( x, y ) = x 2 2 y 2 x + y has its absolute maximum on the shaded tri- angular region D shown in side 2 side 1 side 3 x y 1 1 D 1. at a vertex of D 2. at a critical point inside D 3. on side 1 but not at an end-point cor- rect 4. on side 2 but not at an end-point 5. on side 3 but not at an end-point Explanation:
sikdar (rs52326) – Practice for midterm 2 – perutz – (54175) 25 Now the absolute maximum of f ( x, y ) = x 2 2 y 2 x + y on D occurs either at a critical point of f inside D or at a point on the sides of D . But ∂f ∂x = 2 x 1 , ∂f ∂y = 1 4 y , so f has only one critical point and it occurs at (1 / 2 , 1 / 4), a point inside D since the graph shows that D = braceleftBig ( x, y ) : 0 y x , 0 x 1 bracerightBig .

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