Estimating change in y on a $P=x^2+0.1xy+y^2$, so that we can decrease input of x by 0.5 with P unchanged, given at certain period x=10, y= 8

Question

Full question:

The production $P$ of a company satisfies the equation $P=x^2+0.1xy+y^2$ where $x$ and $y$ are the inputs. At a certain period $x=10$ units and $ y=8$ units. Estimate the change in $y$ that should be made to set up a decrease of $0.5$ in the input $x$ so that the production remains the same.

[Possible Ans: $0.3, 0.61$ or none of the above]

I tried setting it up using implicit differentiation in respect to time [t] but seem unsuccessful, which results me to thinking the answer is "none of the above". Am I correct?

1. If you can tell me a gap in my knowledge based on my writing that'd be great
2. Otherwise can you show me how you'd do this question please?

Answer 1

$P=x^2+\frac{xy}{10}+y^2$

$\frac{dP}{dt}=\frac{d}{dt}(x^2+\frac{xy}{10}+y^2)$

Since production remains same therefore change in it will be $0$

$0=20x\frac{dx}{dt}+x\frac{dy}{dt}+y\frac{dx}{dt}+20y\frac{dy}{dt}$

Plugging values gives

$20*8*\frac{dy}{dt}+8*(-0.5)+10*\frac{dy}{dt}+20*10*(-0.5)=0$

$160*\frac{dy}{dt}-4+10*\frac{dy}{dt}-100=0$

$170\frac{dy}{dt}-104=0$

$\frac{dy}{dt}=\frac{104}{170}=\frac{52}{85}$

Now Since questioner wants to know difference between mine and @Tavish's approach, I have following to tell

See here $P$ depends on two factors $x$ and $y$. . According to question we want to find the rate by which y should change ,when x changes by the rate of -0.5, so that P doesn't change. When we do $dP/dt$ we take in account the rate of change of $P$ when both $x$ and $y$ change so It helps us to get a single equation.

When we do $\partial P/\partial x$ we take in account the rate by which P will be changing when only x changes. When we do $\partial P/\partial y$ we take in account the rate by which P will be changing when only y changes.

You know that

Change in P when both x and y change = Change in P when only x changes + Change in P when only y changes

Change in P when only x changes= rate of change of P when only x changes($\partial P/\partial x$) * change in x

similarly

Change in P when only y changes= rate of change of P when only y changes($\partial P/\partial y$) * change in y

Answer 2

Hint:

When you change $x$ by $-0.5$, the change in $P$ is given by $$ -0.5 \frac{\partial P}{\partial x} $$ Similarly, when you change $y$ by $c$, $P$ changes by $$c \frac{\partial P}{\partial y} $$ You want these changes to cancel out, i.e. $$c\frac{\partial P}{\partial y} -0.5 \frac{\partial P}{\partial x} =0$$ and you can solve for $c$.

Here, e.g. $\frac{\partial P}{\partial x} $ denotes the partial derivative of $P$ w.r.t $x$ at $(10,8)$.