<p>The problem is soluble by taking your formula for the industry sector share <span class="math-container">$f(t)$</span> at time <span class="math-container">$t$</span> and differentiating with respect to <span class="math-container">$t$</span> to determine its gradient at <span class="math-container">$t$</span>, with the maximum where the gradient is zero.  Using the quotient rule and the exponential rule:</p>
<p><span class="math-container">$$\frac{d}{dt}ac^t=ac^t\ln c$$</span></p>
<p>we have:</p>
<p><span class="math-container">$$\frac{df(t)}{dt}=\dfrac{[0.2(1.05^t)\ln1.05(0.2(1.05^t)+0.5(1.02^t)+0.3(1.06^t))]-[0.2(1.05^t)(0.2(1.05^t)\ln1.05+0.5(1.02^t)\ln1.02+0.3(1.06^t)\ln1.06)]}{[0.2(1.05^t)+0.5(1.02^t)+0.3(1.06^t)]^2}$$</span></p>
<p>This may appear daunting, but an immediate simplification is to note that the denominator is positive for all <span class="math-container">$t$</span>, 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 <span class="math-container">$0.2^2(1.05^{2t})\ln1.05$</span> occur within both the first and the second square brackets, so cancel out. The numerator therefore reduces to:</p>
<p><span class="math-container">$$[0.2(1.05^t)\ln1.05(0.5(1.02^t)+0.3(1.06^t)]-[0.2(1.05^t)(0.5(1.02^t)\ln1.02+0.3(1.06^t\ln1.06)]$$</span></p>
<p><span class="math-container">$$=[0.2(0.5)(1.05*1.02)^t(\ln1.05-\ln1.02)]+[0.2(0.3)(1.05*1.06)^t(\ln1.05-\ln1.06)]$$</span></p>
<p><span class="math-container">$$\approx[0.1*1.071^t(0.02899)]-[0.06*1.113^t(0.00948)$$</span></p>
<p><span class="math-container">$$=[0.002899*1.071^t]-[0.000569*1.113^t]$$</span></p>
<p>Setting this equal to zero and dividing through by <span class="math-container">$1.071^t$</span>:</p>
<p><span class="math-container">$$0.002899-(0.000569*(1.0392^t))\approx 0$$</span></p>
<p><span class="math-container">$$1.0392^t\approx5.0949$$</span></p>
<p>which yields <span class="math-container">$t\approx 42$</span>.</p>
<p>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%.</p>

CC BY-SA 4.0
2023-04-02T16:17:42.377Z
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.
VAISHAKH NAIR
CC BY-SA 4.0
2023-04-02T21:18:26.867Z
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.
Adam Bailey
CC BY-SA 4.0
2023-04-03T08:52:35.777Z
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.
VAISHAKH NAIR
CC BY-SA 4.0
2023-04-03T10:09:56.203Z
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.
Adam Bailey
CC BY-SA 4.0
2023-04-02T13:07:50.767Z
2023-04-02T21:32:32.417Z
2023-04-02T21:32:32.417Z
Adam Bailey
<p>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.
<strong>I want to show that Industrial sector's share in  GDP will increase for the first 40 years and then decline.</strong> But I am unable to solve it by hand.</p>
<p>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</p>
<p>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)]</p>
<p>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.</p>

<macroeconomics><mathematical-economics><economic-growth><national-income><growth-accounting>
In a 3 sector economy, calculate maximum possible share of one of the sectors over time if initial sectoral composition and annual growth rate given
VAISHAKH NAIR
# In a 3 sector economy, calculate maximum possible share of one of the sectors over time if initial sectoral composition and annual growth rate given

(Asked by: username_0 on 2023-04-02)

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.

**Tags:** macroeconomics, mathematical-economics, economic-growth, national-income, growth-accounting


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## 1 answers

### Answer by username_1 on 2023-04-02. Score: 1

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%.

> **Comments:**
> **username_0**: 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.  
> **username_1**: 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.  
> **username_0**: 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.  
> **username_1**: 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.  

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