Divided Diamonds : a proportionality & ratio problem

Question

Problem

For stones of the same quality (e.g. the same colour), the monetary value of a diamond is proportional to the square of its mass (i.e. $C\propto M^2$). Find the monetary loss ($\Delta C$) incurred by laser-cutting a diamond worth $C_0$ into two pieces whose masses are in the ratio $a:b$ relative to the original diamond.

Desired Solution

According to the solutions in the back of the textbook.

$\frac{2abC}{(a+b)^2}$

Attempt at Solution

Note: I strongly suspect I am overcomplicating this question, but I have done this question over 3-4 times (building on each other), and would just like another brain to sort out the junk from the jewellery.

(i) Get the new masses ($M_a$ & $M_b$) in terms of the ratio $a:b$.

$M = M\cdot 1 = M\frac{a+b}{a+b} = \frac{M}{a+b} (a+b)= \frac{M \cdot a}{a+b }+ \frac{M\cdot b}{a+b}$

Since $M = M_a + M_b$, by comparing terms we can assert that $M_a = \frac{Ma}{a+b}$ and $M_b = \frac{Mb}{a+b}$.

(ii) Define the old cost and the new cost in terms of $M$, $M_a$, and $M_b$

The old cost, or the initial cost is, $C_0 \propto M^2 \Rightarrow C_0 = kM^2$ , where $k$ is the cost per mass squared.

The new cost, after the diamonds are cut up, is $C_N \propto M_a^2 + M_b^2$ and since we can assume that $k$ doesn't change (e.g. the price of diamonds on the market is stable), then $C_N = k M_a^2 + k M_b^2$.

(iii) Find the difference in cost (i.e. $\Delta C$)

$\Delta C = \text{old cost} - \text{new cost} = kM^2 - k(M_a^2+M_b^2)= k(M^2 - M_a^2-M_b^2)$

Based on what we found in (i):

$\Delta C= k(M^2-(\frac{a\cdot M}{a+b})^2 - (\frac{b\cdot M}{a+b})^2)$ $= k(M^2 - (\frac{a}{a+b})^2 M^2- (\frac{b}{a+b})^2 M^2) = kM^2(1-\frac{a^2+b^2}{(a+b)^2})$

And since $C_0 = kM^2$

$\Delta C = C(1-\frac{a^2+b^2}{(a+b)^2})$

Which is quite different to

$\Delta C = \frac{2abC}{(a+b)^2}$

To all my fellow mathbarians, where didst I misstep?

EDIT: So given the solution, I am stoked. Nevertheless, I wonder whether this is the only way. It is long-winded, but there is no "standard" for how long a solution should be. I didn't expect this to be that long, since I found it in a Year 10 math textbook. I am a university student, and I often overcomplicate things, so I wonder whether that happened here.

Answer 1

$$\Delta C = C(1-\frac{a^2+b^2}{(a+b)^2})=C(\frac{(a+b)^2-(a^2+b^2)}{(a+b)^2})=C(\frac{(a^2+b^2+2ab-(a^2+b^2)}{(a+b)^2})=C(\frac{2ab}{(a+b)^2})=\frac{2abC}{(a+b)^2}$$

Clarification to OP's Further doubts

There can be more easy way,Example

The mass throughout the process is constant Let masses of Pieces be $m_a$ and $m_b$

$m_f=m_a+m_b$

$m_i=m_a+m_b$

As $C$ is proportional to $m^2$ $$\frac{C_f}{C_i}=\frac{m_a^2+m_b^2}{(m_a+m_b)^2}$$ Subtract 1 from both sides(or you can say apply dividendo)

$$\frac{\Delta C}{C_i}=\frac{(m_a^2+m_b^2)-(m_a+m_b)^2}{(m_a+m_b)^2}=\frac{-2m_am_b}{(m_a+m_b)^2}$$

Divide numerator and denominator by $m_b^2$ $$\frac{\Delta C}{C_i}=\frac{-2\frac{m_a}{m_b}}{(\frac{m_a}{m_b}+1)^2}$$

Now it is given that$ \frac{m_a}{m_b}=\frac{a}{b}$

Substituting it in previous equation

$$\frac{\Delta C}{C_i}=\frac{-2\frac{a}{b}}{(\frac{a}{b}+1)^2}=\frac{-2ab}{(a+b)^2}$$

$$\Delta C=\frac{-2C_iab}{(a+b)^2}$$

Negative sign Indicates Decrease in cost