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In an economy, the agriculture, industrial and services sectors have initial shares of 50, 20 and 30 percent respectively in the total GDP. They also subsequently grow at the following constant annual rates for the next 60 years: Agriculture-2 percent; Industry-5 percent, and Services-6 percent per annum respectively. I want to show that Industrial sector's share in GDP will increase for the first 40 years and then decline. But I am unable to solve it by hand.

My understanding is that growth of Industrial sector over time may be represented as (20/100)*(1+0.05)^t = 0.2(1+0.05)^t

Whereas share of Industrial sector will be, growth in industrial sector overt time/ total growth which may be written as f(t) = 0.2(1+0.05)^t/ [(0.2(1+0.05)^t + 0.5(1+0.02)^t + 0.3(1+0.06)^t)]

Maximising the above equation should allow me to find the value of 't' that maximises f(t). However, I do no know how to transform the equation to find the solution. Please let me know how to solve it. Also, is there a faster and more efficient way to approach this question.

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The problem is soluble by taking your formula for the industry sector share $f(t)$ at time $t$ and differentiating with respect to $t$ to determine its gradient at $t$, with the maximum where the gradient is zero. Using the quotient rule and the exponential rule:

$$\frac{d}{dt}ac^t=ac^t\ln c$$

we have:

$$\frac{df(t)}{dt}=\dfrac{[0.2(1.05^t)\ln1.05(0.2(1.05^t)+0.5(1.02^t)+0.3(1.06^t))]-[0.2(1.05^t)(0.2(1.05^t)\ln1.05+0.5(1.02^t)\ln1.02+0.3(1.06^t)\ln1.06)]}{[0.2(1.05^t)+0.5(1.02^t)+0.3(1.06^t)]^2}$$

This may appear daunting, but an immediate simplification is to note that the denominator is positive for all $t$, so if we are only concerned with whether the derivative is positive, negative or zero (and not with its exact value), we can ignore the denominator and focus on the numerator. We may also note that, in the numerator, products equal to $0.2^2(1.05^{2t})\ln1.05$ occur within both the first and the second square brackets, so cancel out. The numerator therefore reduces to:

$$[0.2(1.05^t)\ln1.05(0.5(1.02^t)+0.3(1.06^t)]-[0.2(1.05^t)(0.5(1.02^t)\ln1.02+0.3(1.06^t\ln1.06)]$$

$$=[0.2(0.5)(1.05*1.02)^t(\ln1.05-\ln1.02)]+[0.2(0.3)(1.05*1.06)^t(\ln1.05-\ln1.06)]$$

$$\approx[0.1*1.071^t(0.02899)]-[0.06*1.113^t(0.00948)$$

$$=[0.002899*1.071^t]-[0.000569*1.113^t]$$

Setting this equal to zero and dividing through by $1.071^t$:

$$0.002899-(0.000569*(1.0392^t))\approx 0$$

$$1.0392^t\approx5.0949$$

which yields $t\approx 42$.

Alternatively, the problem can be solved via a spreadsheet (eg Excel) with one row per year and columns for the three sectors, for total GDP, and for industry's share in the total. Using this method and taking the initial position as Year 0, I found that industry's share peaked in Year 42 at 25.167%.

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  • $\begingroup$ Thank you for your response. The things is that I found this question in a multiple choice exam. Hence I was wondering if there is a quicker way to get approximate values. $\endgroup$ Apr 2, 2023 at 16:17
  • $\begingroup$ If you are given options to choose from the problem may be much easier. For example, if given alternative values of $t$ from which to choose that where the industry share is highest, you can just evaluate your formula $f(t)$ at each given value and see which yields the highest share. $\endgroup$ Apr 2, 2023 at 21:18
  • $\begingroup$ I forgot to mention that no calculators are allowed to solve it. Basically the options are, will the industry's share a) keep increasing b) will increase for the first 40 years and then decline, c)will increase for the first 20 years and then decline. I do not know what approach the examiner was expecting. $\endgroup$ Apr 3, 2023 at 8:52
  • $\begingroup$ If calculators were not allowed and there are just those three options, it might help to use the rule of thumb that the doubling period of an annual increase of $n$% is approximately $70/n$. So the sectors double in c 35, 14 and 12 years approximately. Perhaps with some judicious interpolation one could estimate the overall increases after 20, 40 and 60 years. $\endgroup$ Apr 3, 2023 at 10:09

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