Attempt:
Let
$x_l=$ number of barrel of light crude oil.
$x_h=$ number of barrel of heavy crude oil.
Then we should have as objective function
$z=20x_l+15x_h$
And as conditions
s.t. $.4x_l+.32x_h\ge1000000000\\ .2x_l+.4x_h\ge500000\\ .35x_l+.2x_h\ge300,000\\ x_l,x_h\ge0$
Now I am not sure how to add the condition $5\%$ and $8\%$ lost during the refining process.
I though in substracting each of them in each inequality.
Would that be correct?
Someone please help me
prubin has already provided a nice answer, but it seems that you don't get what he says.
See the table.
For light crude oil, add the numbers and you'll get $0.4+0.2+0.35=0.95$ which equals $1-0.05$.
For heavy crude oil, the sum of the numbers is $0.32+0.4+0.2=0.92$ which equals $1-0.08$.
These mean that the loss has already been taken into account.
Hence, we want to minimize $20x_l+15x_h$ under the following conditions :
$$\begin{cases}0.4x_l+0.32x_h\ge 10^6 \\\\ 0.2x_l+0.4x_h\ge 500000 \\\\ 0.35x_l+0.2x_h\ge300000 \\\\ x_l,x_h\ge0\end{cases}$$
Note that you have a typo in the RHS of the first inequality.
Letting $x_l=x,x_h=y$ and considering the graphs, we see that $20x+15y$ is minimized at which the lines $x=0$ and $0.4x+0.32y=10^6$ intersect. (Note that the slope of the line $20x+15y=k$ is $-1.\dot 3$ and that the slope of the line $0.4x+0.32y=10^6$ is $-1.25$.)
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Therefore, the answer is $$20\times 0+15\times\frac{10^6}{0.32}=\color{red}{46875000}$$
Added :
Let $k=20x+15y$ which can be written as $y=-\frac 43x+\frac{k}{15}$. We want to minimize $k$, so that means that we want to minimize the $y$-intercept of the line $y=-\frac 43x+\frac{k}{15}$. Considering the region that the inequalities satisfy, the $y$-intercept is minimized at the red point where the lines $x=0$ and $0.4x+0.32y=10^6$ intersect.
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