What is the minimun & maximum values of objective C.  C= 4x+y  constraints: x-y>=0,  x+3y<=12,  y>=0?
89549
Plot the three boundary lines, shade the correct sides; the answers occur at the vertices of the shaded area (a triangle).  Just plug in these vertices to C = 4x + y and see what the highest and lowest outcome is.
867879
Just plug 0 in the function whenever you see y.\n\n4x + 3(0) + 24 = 0\n4x + 0 + 24 = 0  \n\nThen solve for x.\n\n4x = -24\nx = -6
695016
4x + 3y =12, find y = 4\n4x+3*4=12\n4x+12=12\n4x=0\nx=0
891120
One easy way is to plug in a number for x.  From there, find numbers for y that work with that x.\n\nStart with x = 0 and find a y that works for both inequalities.\nx - 6y < 6\n-x + y < 4\n\nx - 6y < 6\nPlugging in 0 for x, we get\n-6y < 6\ny > -1\n\n-x + y < 4\nPlugging in 0 for x, we get\ny < 4\n\nWhen x is 0, y has to be greater than -1 and less than 4.\nThat gives us four solution points already.\n(0,0), (0,1), (0,2), (0,3)
753539
Since f(x) is a parabola that opens upward, you need the minimum value, which is the y-coordinate of the vertex.  The x-coordinate is -(-12)/(2•3) = 2.  So the minimum y is at 3(2)² - 12(2) + 13 = 1, and your range is y >= 1.\n\nFor the second part, just solve 3x² - 12x + 13 = 28.\n\n3x² - 12x - 15 = 0\nx² - 4x - 5 = 0\n(x-5)(x+1) = 0\nx = 5 or x = -1.
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.
604546
your equation is a little weird - i believe you mean:\nh(x)=-0.03x^2+3x\nin any case, finding the "maximum" or "minimum" of anything with a given equation is simply finding the derivative of the equation, finding "x" when that equals 0, and finding the height or whatever when x=that number.  for example, the equation above is a quadratic that looks like an arc if you graph it.  finding the derivative:\n\nh(x)=-0.03x^2+3x\nh'(x)=-(2)0.03x+3\nh'(x)=-0.06x+3\n\nnow let h'(x)=0.  this means, by looking at the graph, when the height of the arc is no longer increasing or decreasing, or has reached its max or min.  find x when h'(x)=0:\n\nh'(x)=-0.06x+3=0\n-0.06x+3=0\n-0.06x=-3\nx=3/(0.06)\nx=50\n\ntherefore the maximum height of the arch occurs when x=50.  to find the height at this point, just plug it into the given equation:\n\nh(50)=-0.03(50)^2+3(50)\nh(50)=-75+150\nh(50)=75\n\ntherefore the maximum height of the arch is 75 meters.  hope this helped!
863968
-3x^2+24x-49\ndy/dx=-6x +24\n-6x+24 =0, so x=4 \n\nSo the max occurs when x=4. ( We know it is a max because the coefficient of x^2 is negative.)\nso the max is -3(4^2) +2*4-49 = -48 +8 -49= -89\nThis means that the range (possible values of y) is all the real numbers =< -89 . Because -89 is the max value. In interval notation this is - infinity <y<= -89.\n\nThe graph is a parabola which is increasing upward until it hits the max when x = 4 and then decreases forever\nThus f(x) is increasing over the interval - infinity <x<4  and it is decreasing over the interval 4<x< + infinity. At x =4, f(x) is neither increasing or decreasing.\n\nThe vertex is located at (4, -89)\n\nSince f(x) is never positive, it never crosses the x-axis and thus has no real roots. The two roots are imaginary.\n\nI think that covers all your questions.
1077864
Assuming D is z=24x+8y\nThe max with those numbers would be C with a Z value of 24.5 times 25 or 612.5
764433
Finding max's and mins is done thru the first derivitive and setting it to 0, so\n\nf(x)= sinx - cosx\nf1(x)= cosx + sinx\n0=cosx + sinx\n-cosx = sin x\nthis occurs when x= (3pi)/4\nmax is root 2, as was stated.\n\nThe answer that used to be directly above is wrong.
700951
4x + 6 is not quite right\nwatch the order of operations and distributive properties\n\nrepost your answer and I'll tell you if you got it\n\n(you didn't think I'd just give up the answer, did you ?)
885850
y'=-10x+2\nsetting itto 0\nx=1/5\ny''=-10 negative so a max at x=1/5\nthe maximum value=-5(1/5)^2+2/5-4\n=-1/5+2/5-4\n=-19/5
147089
y intercept: (0,4)\nx intercept: (3,0)
664485
nope \n5x+5y\n\nthe 4y and -4y cancel each other out
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!
679397
y = 0.5x-4 \n\nor similarly\n\nx= 2y +8
706557
4x+4>(with line under arrow) 4 is what you called as a linear inequalities. To graph this inequality, convert this inequality into an equation so it will become 4x+4=4. Solve this using  the intercept method. The next step is to plot the x-intercept and y-intercept in the cartesian plane. Your equation can be plot because you have only one variable so you can only get x-intercept but you cannot get the y-intercept. for this, when there are no y variable in the inequality, the value of y-intercept is zero and the value of x-intercept is 0 so the plot of this inequality is in the coordinates (0,0)