Body Building Month

Let $x$ be amount of protein bars $A$ and $y$ be amount of protein bar $B$ .

From the information in the question, we can set up system of linear inequalities as following:

$x \geq 0$

$y \geq 0$

$200x + 240y \geq 20,000$

$15x + 30y \geq 1,800$

$30x + 20y \geq 2,200$

Then when drawing the graphs for these inequalities, we will obtain the red line for the calories linear equation $5x + 6y = 500$ , blue for protein equation $x + 2y =120$ , and green for carbohydrate equation $3x + 2y= 220$ , with yellow region as the viable coordinates meeting all constraints:

According to the linear programming, the optimal solution to the cost equation $Z = 3x + 4y$ will lie upon one of the vertices of the yellow polygon. Therefore, there are $4$ intersecting points to consider: $(0 , 110), (40 , 50), (70 , 25), (120 , 0)$

Plugging in the coordinate values for $Z$ , we will get:

$Z(0 , 110) = 4\times 110= 440$

$Z(40 , 50) = (3\times 40) + (4\times 50) = 320$

$Z(70 , 25) = (3\times 70) + (4\times 25) = 310$

$Z(120 , 0) = 3\times 120 = 360$

As a result, the bodybuilder can spend $\boxed{310}$ dollars, which is the least possible amount of money, to purchase the protein bars and meet his one-month requirements.

Another take on this, the above information gives 3 equations (A is bar A, B is bar B): $15A+30B=1800\quad (1)$ $30A+20B=2200\quad (2)$ $200A+240B=20000\quad (3)$ It's ineffective to solve a system of three equations with only two variables, but you can group these in 3 sets of two and solve each of those. $(1)\space and \space (2): A=50, B=35$ $(1)\space and \space (3): A=70, B=25$ $(2)\space and \space (3): A=40, B=50$ Now check each pairing with the third equation to make sure it works. (1) and (2) do not give enough calories, so that one is out. (1) and (3) give 2600g of carbohydrates, and (2) and (3) give 2100g of protein. Those both work. $(1)\space and \space (3)\space costs \space \$\color{#D61F06}{310}$ $(2)\space and \space (3)\space costs \space \$320$ The least price wins.

Let us presume we don't have a method of graphically displaying the inequalities (as I often don't while scratching my daily solving itch). Also, let 1 protein = P, 1 carbohydrate = C, 1 calorie = E, and 1 dollar = D. From the problem,

A=15P, 30C, 200E, 3D B=30P, 20C, 240E, 4D

On a nutrition per D basis, A=5P, 10C, $66.\dot { 6 }$ E B=7.5P, 5C, 60E

From this we can see that bar B provides the best P per D, and bar A provides the bes C and E per D.

To solve from here, we need to bring one of P, C, or E to the required value using only one type of bar, then add the other type of bars to increase the other variables whilst holding the first constant by removing the first type of bar. As bar A is more cost effective in both C and E, we will first use bar B to minimise the cost of P.

$1800\div 30=x\\ x=60$

60B=1800P, 1200C, 14400E

From the problem, we know that if we remove one bar of B, we must replace it with two bars of A to maintain the same total P. This will increase C by 40 E by 60, and D by 2. We need to increase C by 1000, so we will need $1000\div 40=25$ exchanges to reach the target C value. This will result in 50 A and 35 B.

50A+35B=1800P, 2200C, 18400E

We still require 1600C to satisfy the question requirements. Using just bar A, this will cost $1600\div 200\times 3=24D$ . Using exchanges, it will cost $1600\div 160\times 2=20D$ . (We have already established that Bar B is more expensive per D than bar A, and therefore is excluded from consideration). Therefore, we require $1600\div 160=10$ exchanges, resulting in 70A and 25 B.

70A+35B=1800P, 2600C, 20000E, 310D, or $310

Seriously, you need to fix the wording of this problem to make clear that "at least" 2,200 g of carbohydrates is required. The way it reads now, it can be interpreted to read, "exactly 2,200 g of carbohydrates", which which case the minimum monthly cost is $320, not $310.

I solved using Minimize on Mathematica.

https://reference.wolfram.com/language/tutorial/ConstrainedOptimizationLinearProgramming.html