2x50 100x1 4x25
ok im not sure exactly what you mean but here goes 10x10 20x5 4x25 100x1 1x10to the second power
one fifth of 100 coins is 100X1/5= 20
firstly, 900mm is 0.9m. 100x1 is 100, x0.9 is 90m3
Infinitely many. Suppose the area of the rectangle is 100. We could create rectangles of different areas: 100x1 50x2 25x4 20x5 10x10 However, the side lengths need not be integers, which is why we can create infinitely many rectangles. Generally, if A is the area of the rectangle, and L, L/A are its dimensions, then the amount 2(L + (L/A)) can range from a given amount (min. occurs at L = sqrt(A), perimeter = 4sqrt(A)) to infinity.
Because you haven't inquired if it is simple or compund interest, I'll just put up both and you choose the right one. Also, you might use an alternate formula to me (I'm Australian, and besides there are heaps of variations): Simple interest I= PRT/100 =(100 000x5x10)/100 =5 000 000/100 =$5 000 Total=100 000+5 000 =$105 000 Compound interest A=P(1+r)^n =100 000(1+(5/100x1))^10 =100 000x1.6289 =$162889.46
This is a classic max min problem using Lagrange multipliers. We do this when we have a function of several variables that we want to maximize or minimize and a constraint. Now the equation for surface area of a box is the areas of all the sides added together Total Surface Area = 2(Areahxw) + 2(Areahxl) + 2(Areawxl) using x,y and z we have f(x,y,z) = 2(xy+xz+yz) we need the gradient of that which is <(partial derivative wrt x,), partial derivative wrt y,(partial derivative wrt z)> where wrt means with respect to and i am using <> for a vector. 2 = ∇f(x,y,z) remember that the gradient is a vector and is normal to the curve. now let g be the other function, the one we have constraints on. In this case x+y+z=150 so that sum is constant and we write g(x,y,z)=x+y+z-150=0 we know that ∇f(x,y,z)=2 now the gradient, ∇g(x,y,z) is <1,1,1> since the partial derivative of x+y+z-150 wrt to x is 1 etc. so we have 2=<1,1,1>λ in order for these to be equal, the components must be equal =λ/2<1,1,1> now we have several equations y+z=λ/2 x+z=λ/2 x+y=λ/2 x+y+z=150 so we have y+z=x+z so x=y we also have y+x=y+z so x=z this means x=z=y and using the last equation we have 3x=150 or x=50 so y=50 and z=50 the rectangular box must be 50x50x50 the surface area is 2(2500+2500+2500) check a few other values of x, y and z and see what surface area you get. Perhaps, 100x49x1? we've only got one solution we might be tempted to assume that these are the dimensions that will give the smallest surface area.. The method of Lagrange Multipliers will give a set of points that will either maximize or minimize a given function subject to the constraint, provided there actually are minimums or maximums. So the points 100,49,1 give us info if we found a max or a min 50x 50 x 50 box has a surface area of 15000 the 100x49x1 box has a surface area of 2(100x49+100x1+49x1)=10098 which is less than 15000 so we found a max not a min! since we have x+y+z=150 if one of them gets very small the others have to get very large to compensate, They cannot all get very big or they won't work with our constraint. So the fact that we have a max them seems to make sense. In others, take very small x and y say .01 each then z=149.98 we get a surface area of about 6 or so. (much less than 15000) So now we wonder, do we have a minim value? We looked at .01,.01,149,98 let's take that one more step. x=y=.001 z=150-.002=149.998 surface area is around 1/3 so it looks like we could keep making x and y smaller and smaller and the surface area would continue to shrink. So no matter what values you give me for x and y and z, I can find a box with smaller surface area. remember that a zero derivative may mean a max or min, or may be a saddle point, but a max or min always has a zero derivative. For constrained optimization problems, look and see if the constraint is bounded, for example a circle. The graph is x+y+x-150=0 is a plane in 3 space and is not bounded. This tells us about the existence of max and min.