Question

How do you find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum?

Answer 1

This question is from the chapter of calculus. In this question, first we take two variables that will be x and y as it asks for positive numbers. After that, we will multiply the variables that will be equal to zero. After that we will add both of them in which one of them is multiplied by 10. The addition should be minimum. So the first derivative of that addition will be zero. And the second derivative should be negative at the point where the first derivative is zero. After that, we will solve both the equations and get the value of x and y.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to find the two positive numbers whose product is 750 and also the sum of one and 10 times the other is minimum.
So, let us take the two positive numbers as x and y.
As it is given that product of them is 750. So, we can write
$x\times y=750$
$\Rightarrow y=\dfrac{750}{x}.............(1)$
And also it is given that the sum of one and 10 times the other (where the sum will be x+10y) is minimum.
So, we can write
$\dfrac{d}{dx}\left( x+10y \right)=0$
Using the value of y from the equation 1, we can write the above equation as
$\Rightarrow \dfrac{d}{dx}\left( x+10\left( \dfrac{750}{x} \right) \right)=0$
$\Rightarrow \dfrac{d}{dx}\left( x+\dfrac{7500}{x} \right)=0$
$\Rightarrow \dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( \dfrac{7500}{x} \right)=0$
Using the differentiation formulas, we get
$\Rightarrow 1-\dfrac{7500}{{{x}^{2}}}=0$
$\Rightarrow 1=\dfrac{7500}{{{x}^{2}}}$
$\Rightarrow {{x}^{2}}=7500$
Squaring root both sides, we get
$\Rightarrow x=\sqrt{7500}$
The prime factorization of 7500 will be $2\times 2\times 3\times 5\times 5\times 5\times 5$
So, we can write
$\Rightarrow x=\sqrt{2\times 2\times 3\times 5\times 5\times 5\times 5}=2\times \sqrt{3}\times 5\times 5=50\sqrt{3}$
The second derivative of the sum should be positive for minimum value. So, we will check the value of the second derivative for further solving.
$\Rightarrow \dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( x+\dfrac{7500}{x} \right)=\dfrac{d}{dx}\left( 1-\dfrac{7500}{{{x}^{2}}} \right)=\dfrac{2\times 7500}{{{x}^{3}}}$
Hence, the value of the above equation at x=$50\sqrt{3}$ will be positive. So, the sum is minimum.
Now, we will do further solving.
In the equation (1), putting the value of x as $50\sqrt{3}$, we get
$\Rightarrow y=\dfrac{750}{50\sqrt{3}}=\dfrac{50\times 15}{50\sqrt{3}}=\dfrac{50\times 5\times 3}{50\sqrt{3}}=\dfrac{50\times 5\times \sqrt{3}\times \sqrt{3}}{50\sqrt{3}}=5\sqrt{3}$

Hence, the two positive numbers are $50\sqrt{3}$ and $5\sqrt{3}$.

Note: For solving this type of question, we should know about maxima and minima. If something is given as maximum or minimum, then the first derivative of both will be zero and the second derivative for maximum will be negative and the second derivative for minimum will be positive at the point where the first derivative was zero. We should have a better knowledge in differentiation so that we can solve this type of question easily. Don’t forget the formula of differentiation like $\frac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ .