Need with this asap! analysis question involving math

  2000 of Type A; 3000 of Type B; profit of $74 million

You have enough of these similar questions that it seems like it would pay to find a generic solution.

Let revenue (R) be ...

  R = ax + by

and cost (C) be ...

  C = cx² +dxy +ey² +fx +gy +h

Then profit (P) is the difference between revenue and cost ...

  P = R - C = ax + by - (cx² +dxy +ey² +fx +gy +h)

  P = -cx² -dxy -ey² +(a-f)x +(b-g)y -h

The maximum is found where the partial derivatives with respect to x and y are zero.

  ∂P/∂x = -2cx -dy +(a-f) = 0

  ∂P/∂y = -dx -2ey +(b-g) = 0

These two equations can be written in standard form as ...

  2cx +dy = a-f

  dx +2ey = b-g

Then Cramer's rule gives the generic solution to this linear system as ...

  x = (d(b-g) -2e(a-f))/(d² -4ce)

  y = (d(a-f) -2c(b-g)/(d² -4ce)

Using your numbers ...

  {a, b, c, d, e, f, g, h} = {4, 6, 1, -3, 9, 9, -42, -7}

we get ...

  x = (-3(6-(-42)) -2·9(4-9))/((-3)² -4·1·9) = (-3·48 -18(-5))/-27 = -54/-27 = 2

  y = (-3(-5) -2·1·48)/-27 = -81/-27 = 3

And the profit is ...

  P = -1(2²) +3(2)(3) -9(3²) -5(2) +48(3) +7 = -4 +18 -81 -10 +144 +7 = 74

Sale of 2000 type A units and 3000 type B units will yield the maximum profit of $74 million.

black/brown

step-by-step explanation:

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55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.