dy/dx= 2x therefore d2ydx2= 2 as this is positive we can tell that it is the minimum value in a curve
Chat with our AI personalities
4x + 3y = -5 (Eq 1); 3x + 5y = -12 (Eq 2) Eliminate x Eq 1 times 3: 12x + 9y = -15 Eq 2 times 4: 12x +20y = -48 Subtract Eq 1 from Eq 2 11y = -48-(-15) ie 11y = -33 y = -3 Substitute in Eq 1: 4x - 9 = -5, ie 4x = 4 so x = 1 Check in Eq 2: (3 x 1) + (5 x -3) = 3 - 15 = -12. QED
x + 7y = 39 ....... Eq.13x - 2y = 2 ....... Eq.2multiply Eq.1 with (-3) and so Eq.1 will be :-3x - 21y = -1173x - 2y = 2_________________ by Elimination method ( Eq.1 + Eq.2 ) the result will we :- 21y - 2y = - 115-23y = -115Y = 5to find the value of x, put the value of y in Eq.1 :x + 7y = 39x = 39 - 7yx = 39 - 7(5)X = 4___________________________________________________________you can solve it in another way by using substitution method:x + 7y = 39 ....... Eq.1 ,, you can write it in another form, x = 39 - 7y3x - 2y = 2 ....... Eq.2Put the value of Eq.1 on Eq.23(39 - 7y) - 2y = 2117 - 21y - 2y = 2117 - 23y = 223y = 115 ............ Y = 5to find the value of X , put the value of Y in Eq.1x = 39 - 7yx = 39 - 7(5)x = 39 - 35 .............. X = 4and so in both ways....you got the same answer choose the easiest way for you :)good luck my friend...
16
Two contour lines can intersect. A perfect example is a Lagrange Multiplier which is encountered in Calculus III. We are given a function that has restraints (side conditions). An optimization engineer working for a box factory might be asked to find the maximum volume of a cardboard box given the restraint that it has a surface area of 1500 cm2 and a total edge length of 200 cm.We are seeking the extreme values of f(x,y,z) that lie on the one of the level curves (c) of g(x,y,z) and h(x,y,z). These occur at a point P(x0,y0,z0) where you can find the highest level surfaces (k) of f(x,y,z) that are intersected by the level curves (c) of g(x,y,z) and h(x,y,z). These intersections occur when they just barely touch one another. Meaning they have a common tangent line. Further, their normal lines are the same, implying that their gradient vectors ∇f, ∇g, ∇h are parallel.∇f = λ∇g + μ∇h. This works if ∇g and ∇h ≠ 0.Eq. 1 f: V=xyzEq. 2 g: 1500=2(xy)2+2(xz)2+2(yz)2Eq. 3 h: 200=√x2+y2+z2∇f =(yz,xz,xy)∇g = (4xy2+4xz2,4x2y+4yz2,4x2z+4y2z)∇h = (x/√x2+y2+z2, y/√x2+y2+z2, z/√x2+y2+z2)Eq. 4 yz= λ(4xy2+4xz2) + μ(x/√x2+y2+z2)Eq. 5 xz= λ(4x2y+4yz2) + μ(y/√x2+y2+z2)Eq. 6 xy= λ(4x2z+4y2z) + μ(z/√x2+y2+z2)We have 6 equations and 6 unknowns (x,y,z,λ,μ and V). We will have to use back substitution to solve.
16x - 2y = 74 (Eq 1) 2x - 2y = 4 (Eq 2) from Eq 2, 2y = 2x - 4 Substitute in Eq 1 16x - (2x - 4) =74 ie 16x - 2x + 4 = 74 ie 14x = 70 ie x = 5 2y = 2x - 4 = 10 - 4 = 6 y =3 Check 16 x 5 - 2 x 3 = 80 - 6 = 84 QED