What is the minimun & maximum values of objective C. C= 3x+y constraints: 3x+2y>=12, -x+y<=1, x<=4?
89202
Hi again, just graph the three boundary lines and shade the correct sides.  This will give you a shaded triangle with vertices (2,3), (4,5), and (4,0)  [I think; I don't have graph paper handy].  The maximum and minimum values must occur at one of these 3 sets of coordinates.\n\nfor (2,3), 3x + y = 9\nfor (4,5), 3x + y = 17\nfor (4,0), 3x + y = 12\n\nSo 9 is the minimum value and 17 is the maximum.\n\nCheck it out yourself to be sure the graphs really intersected where I thought they did!
321913
the answer is letter "C". coz 3(3x+y)=9x+3y.use the distributive property.now do your homework lazy kid!!! j/k!!! lmao!!! =D haha! kidding! hope i helped you somehow!
71941
In the Formula Bar, type:\n=MAX(First_cell:Last_cell)\n\nFor example:\n=MAX(A1:C255)
718048
x + y = 3\n\nSolve for a variable.  I chose X.\nx+y = 3\n -y    -y\nx = 3 - y\n\nSubstitute the value of X in the first equation.\n2(3-y) + y = 4\nDistribute.\n6-2y+y = 4\nAdd like terms.\n6-y = 4\n-6    -6\n-y = -2\nmultiply by -1.\ny=2\n\nSince you know Y = 2, substitute that back into the second equation that you solved for X with.\nx + y = 3\nX + 2 = 3\n  -2    -2\nX = 1\n\nX = 1\nY = 2
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)
143223
a) (6x-7y)(6x+7y)\nb) y(4xy-7)\nc) (x-8)(x+3)
593080
*** pseudocode ****\n\n\n$max = $array[0];\n$min = $array[0];\n\nfor( $x=1; $x <= $size_of_array; $x++ )\n {\n     if($array[$x] < $min)\n       {\n          $min = $array[$x];\n       }\n\n     if($array[$x] > $max)\n       {\n          $max = $array[$x];\n       }\n\n}\n\n\n\n\nhave fun\n\n\nt
561113
y= -4x+8x-1 = y=4x-1\nThere is no vertes as this is the equation of a straight line.\n\nAssuming that you made a typo, and this is what you meant, y= -4x^2+8x-1, then\n\ny = -(4x^2-8x+1)\ny = -[(2x-2)^2 - 4 + 1]\ny = -[(2x-2)^2 - 3]\ny = 3 - (2x-2)^2\n\nSo the vertex is (1,3) and it is a maximum point.
787980
i believe this is a continuation of the previous prob.\n\nit can be solved graphically. \ndraw the lines x + y = 2,  y - x = 2 and y = -2 on the cartesian plane. the feasible reason will be the area enclosed by the triangle formed by these 3 lines. the vertices being (0,2) , (4, -2) and (-4, -2). it can be proved that the max value of 2x + 3y will be one of these vertices. \n\nsubstituting, u can see the maximum occurs at (0,2) whence 2x + 3y = 6
1146893
x^2+mx+1/x^2+x+1<3\n\ncase 1 : for x>0\n=> (x+1/x)^2 +(m+1)x <4\n=> (m+1)x < 4- (x+1/x)^2\n\nNow, we find the maximum value of the RHS:\nthe minimum value of (x+1/x)^2 is 2 when x=1 and for all other values of x it is >2 \nThis makes max(RHS)=0 or m+1 <0 \nor m<-1\n\ncase 2: x<0\nthe same argument as above(but x=-1)  leads us to m+1 >0 \nor m>-1\n\ncase 3: x=0\nThe given expression is undefined \n\nI guess its a home work problem, which indeed makes you lazy!
705864
Alright, don't listen to the above post...\n\nGrade 12 here, answering questions all day for mere fun.\n\n12)\ny - 3 = -1/3 (x + 4)\n\nEXPLAINED\n\nadd 3 to both sides\ny = -1/3 (x + 4) + 3\n\nexpand 1/3(x + 4) \n= x * 1/3 + 4 * -1/3\n= x/3 + 4/3\ny = -1/3x - 4/3 + 3\n\nTo add -4/3 and 3, use 3 = 9/3\ny = -1/3x - 4/3 + 9/3\ny = -1/3x + 5/3\n\nTrue, my bad. I originally added 4/3 instread of -4/3\n\n\n16)\n(-2, -1) (4, -4)\n\nm = Dy / Dx\nm = ((-1) - (-4)) / ((-2) - (4))\nm = 3 / -6\nm = -1/2\n\ny = mx + b\n(-1) = -1/2(-2) + b\n-1 = 1 + b\n-1 - 1 = b\n-2 = b\n\ny = -1/2x - 2\n\nCHECK\ny = -1/2(4) - 2\ny = -2 - 2\ny = -4\nTherefore it works.\n\n18)\n(0, 0) (2, 3)\n\nm = Dy / Dx\nm = ((3) - (0)) / ((2) - (0))\nm = 3/2\n\nb = 0\n\ny = mx + b\ny = (3/2)x + 0\ny = (3/2)x\n\nNote, the lower post copied my answers
939650
x+y=-3\nx=-y-3\nx -3  -2  -1  0  1  2  3 \ny  0  -1  -2 -3-4-5 -6\n\nx^2+2y^2-12y-18=0\nwhen y=-1\nx^2+2+12-18=0\nx^2-4=0\nx^2=4\nx=+/-2\nso (-2,-1),(2,-1)\nwhen y=3\nx^2+18-36-18=0\nx^2-36=0\nx=+/-6\nso(-6,3),(6,3)\n\npoints of intersection\n(-2,-1) and (-6,3)
147089
y intercept: (0,4)\nx intercept: (3,0)
763532
put in y=3x-8 in x+y=4 like this\n\nx+(3x-8)=4\n\n4x-8=4\n   +8  +8\n\n4x=12\n/4    /4\n\nx= 3
613986
=IF(E12=0,0,\nIF(AND(E12>0,E12<4),\n5.9,IF(E12>=4,12.9,\n"negative value")))\n\nAll on one line!\n\nEdit:  The >= (greater than or equal to) is required else the value of exactly 4 would return "negative value"
1028088
x=y+150\n\nYou will need another statement with x & y or just insert a value for x or y and resolve the question above.
720107
4x -12y =5...........(1)\n-x +3y = -1..........(2)\n\nMultiplying eq.(2) by 4 and add with eq.(1)\n(4x -12y =5) + (-4x +12y = -4)=\n0 = 1\nwhich is invalid.