Answer:
The metallurgist should use 3 ounces of the 30 % copper alloy and 9 ounces of the 50 % copper alloy to make 12 ounces of 45 % copper alloy.
Step-by-step explanation:
The ounce is a mass unit, as we notice that the metallurgist wants to make 12 ounces of an alloy containing 45 % copper by mixing two metal with different copper proportions. We can use the following two equations:
Alloys
[tex]m_{R} = m_{A}+m_{B}[/tex] (Eq. 1)
Copper
[tex]r_{R}\cdot m_{R} = r_{A}\cdot m_{A}+r_{B}\cdot m_{B}[/tex] (Eq. 2)
Where:
[tex]m_{A}[/tex] - Mass of the 30 % copper alloy, measured in ounces.
[tex]m_{B}[/tex] - Mass of the 50 % copper alloy, measured in ounces.
[tex]m_{R}[/tex] - Mass of the 45 % copper alloy, measured in ounces.
[tex]r_{A}[/tex] - Proportion of copper in the 30 % copper alloy, dimensionless.
[tex]r_{B}[/tex] - Proportion of copper in the 50 % copper alloy, dimensionless.
[tex]r_{R}[/tex] - Proportion of copper in the 45 % copper alloy, dimensionless.
Now, the mass of the 50 % copper alloy is cleared in Eq. 1 and eliminated in Eq. 2:
[tex]r_{R}\cdot m_{R} = r_{A}\cdot m_{A} + r_{B}\cdot (m_{R}-m_{A})[/tex]
[tex](r_{R}-r_{B})\cdot m_{R} = (r_{A}-r_{B})\cdot m_{A}[/tex]
And we clear and calculate the mass of the 30 % copper alloy:
[tex]m_{A} = m_{R}\cdot \left(\frac{r_{R}-r_{B}}{r_{R}-r_{A}} \right)[/tex]
If we know that [tex]m_{R} = 12\,oz[/tex], [tex]r_{R} = 0.45[/tex], [tex]r_{A} = 0.30[/tex] and [tex]r_{B} = 0.50[/tex], the mass of the 30 % copper alloy:
[tex]m_{A} = (12\,oz)\cdot \left(\frac{0.45-0.50}{0.30-0.50} \right)[/tex]
[tex]m_{A} = 3\,oz[/tex]
And the mass of the 50 % copper alloy is:
[tex]m_{B} = m_{R}-m_{A}[/tex]
[tex]m_{B} = 12\,oz-3\,oz[/tex]
[tex]m_{B} = 9\,oz[/tex]
The metallurgist should use 3 ounces of the 30 % copper alloy and 9 ounces of the 50 % copper alloy to make 12 ounces of 45 % copper alloy.
