Solution to "Mauricio's Uterus Problem"


The basic problem here is that the label does not specify exact amounts of the ingredients, so we have to see if we can find them, or at least a range, with the given information.

So we know that the 3 extracts, which we will call r, g, and s, for red raspberry, ginger juice, and spirulina, respectively, add together to equal 90 mg. In other words,

r+g+s=90.

The equivalent raw amounts are, using the information on the label, 2r, 10g, and s. These 3 add together, according to the label, to be an equivalent raw amount of 320 mg. So

2r+10g+s=320.

These are the only equations we can get from the information. Since there are more unknowns than equations we cannot find an exact answer. But we have conditions that r>0, g>0, and s>0, which may be able to give us a range with which we may be able to answer the problem.

To find all solutions to a system of equations in which there are fewer equations than unknowns, first we want to find just one solution to the equations, and then add a scalar multiple of one solution to the corresponding homogeneous system (set the constants to 0). So our general solution (without respect for condition of positive amounts) will be of the form

\left( \begin{array}{c}r\\ g\\ s\end{array}\right) = \left( \begin{array}{c}r_p\\ g_p\\ s_p\end{array}\right) + c \left( \begin{array}{c}r_h\\ g_h\\ s_h\end{array}\right)

where (r_p, g_p, s_p) is one particular solution, c is any real number, and (r_h, g_h, s_h) is one nontrivial solution to the homogeneous equations.

So our two equations are

r+g+s=90,\text{ and}

2r+10g+s=320.

Subtracting equation 1 from equation 2, we get

r+9g=230.

Remember, we only need one particular solution right now, so now we can just pick an r<230, solve for a corresponding g, then put both back in the original equation to get an s. I chose

r_{p}=50

g_{p}=20

s_p=20

as my particular solution, but you will get to the same final answer with any particular point that works in these equations. Now, we need a (r_h, g_h, s_h), so our homogeneous equations are

r+g+s=0,

2r+10g+s=0.

This time, subtracting the first equation from the second equation gives us

r+9g=0.

Therefore,

r=-9g.

Letting g_h=1 then r_h=-9 and s_h=8. So we have our one nontrivial solution to the homogeneous system. Then, as we stated earlier, the general form of all solutions to these 2 equations is

\left( \begin{array}{c}r\\ g\\ s\end{array}\right) = \left( \begin{array}{c}r_p\\ g_p\\ s_p\end{array}\right) + c \left( \begin{array}{c}r_h\\ g_h\\ s_h\end{array}\right) = \left( \begin{array}{c}50\\ 20\\ 20\end{array}\right) + c \left( \begin{array}{c}-9\\ 1\\ 8\end{array}\right).

This will give us all possible solutions to the system where c is any real number. For instance, letting c=1 results in

r=50+1(-9)=41

g=20+1(1)=21

s=20+1(8)=28.

Then (41, 21, 28) is also a solution. Let's check that:

41+21+28=90

2(41)+10(21)+28=320.

Okay, so that does indeed work to find other solutions. The only thing left then is to impose the conditions r>0, g>0, and s>0. To do this, we will find what values of c will cause r=0, g=0, or s=0 and it will give us a maximum and/or minimum value for c.

So if c is positive, it will at some point cause r=50-9c=0. And that happens at c=50/9. Then

c<50/9

since c>50/9 will push r into the negatives. Similarly, we can see that if c is negative, it will at some point cause s=20+8c=0. And this happens at c=-5/2. So if c< -5/2 then s will be pushed into the negatives, which we cannot have, therefore

c> -5/2.

Then our upper bound and lower bounds for c give us

 -5/2 < c < 50/9.

Then putting these upper and lower bounds into our original equations, we get a necessary range for r, g, and s:

\left( \begin{array}{c}r_{\text{max}}\\ g_{\text{min}}\\ s_{\text{min}}\end{array}\right) = \left( \begin{array}{c}50\\ 20\\ 20\end{array}\right) + (-5/2) \left( \begin{array}{c}-9\\ 1\\ 8\end{array}\right) = \left( \begin{array}{c}145/2\\ 35/2\\ 0\end{array}\right) = \left( \begin{array}{c}72.5\\ 17.5\\ 0\end{array}\right)

and

\left( \begin{array}{c}r_{\text{min}}\\ g_{\text{max}}\\ s_{\text{max}}\end{array}\right) = \left( \begin{array}{c}50\\ 20\\ 20\end{array}\right) + (50/9) \left( \begin{array}{c}-9\\ 1\\ 8\end{array}\right) = \left( \begin{array}{c}0\\ 230/9\\ 580/9\end{array}\right) \approx \left( \begin{array}{c}0\\ 25.56\\ 64.44\end{array}\right).


Then

0<r<72.5

17.5<g<25.56

0<s<64.44.

Remember, these are ranges for the extracts. To answer the question, we need to know raw amount ranges, so simply converting these with 2r, 10g, and s gives us raw ranges of

0<r_{\text{raw}}<145

175<g_{\text{raw}}<256

0<s_{\text{raw}}<65.

So the maximum amount of raw ginger Mauricio will get from the vitamin is 256 mg, and he would have to have more than 270 mg to explode; therefore he will not explode. But the maximum amount of raw red raspberry he can get from the vitamin is 145 mg, and he needs at least 150 mg to prevent his uterus from hurting the next day, so unfortunately his uterus will indeed hurt the next day.