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This is one of our many quizzes on Optimization. Click on this to access more such competitive quizzes on Algebra.

Optimization Quiz 2

mathematics

 Given that optimisation is based on the selection of the best element amongst a group of various other alternatives, which of the following more closely describes the definition of optimisation?

 Identify the best option which appropriately describes optimisation.

Option A: The instantaneous rate is the derivative of a function which is one of the steps involved in optimisation but does not provide an optimal solution. Option B: Optimisation minimizes or maximizes a function relative to some arbitrary set such that it represents a range of choices in a certain situation. Option C: Solution elimination is not optimisation since there is no mathematical approach involved. Optimisation has a mathematical approach. Option D: The principal branch is usually applied in multivariable functions that are present in the complex plane and are useful for complex analysis.

 Optimisation has certain constraints attached to it. Identify the necessary conditions for problems based on optimization.

 Identify the best option which appropriately mentions the necessary conditions for problems based on optimisation. 

Option A: The Lagrange multiplier method is used to calculate the maximum or the minimum of a particular function f(x) which has an equality constraint such that $g\left( x \right)=0$. It can be evaluated using the Lagrangian function $\Rightarrow \text{Lagrange}\left( x,\text{ Lambda} \right)=f\left( x \right)-\text{Lambda}\times \text{g}\left( x \right)$ Option B: Critical points of a function where the function is either not differentiable or has its derivative = 0 or it is not holomorphic. Option C: Stationary points such as the first derivative of the function is evaluated to 0 is one of the critical points of a function. Since all of these are necessary conditions in various problems based on optimisation, option D is correct.

 Given that two positive numbers $a$ and $b$ have a sum of $450$ such that the product $ab$ is a maximum. Identify the two numbers and the product. 

 Using the information given in the question, set up $2$ constraints and express each in terms of $1$ variable. Calculate the $1\text{st}$ and $2\text{nd}$ derivative of the function that has a maximum and solve for each variable. 

$1\text{st}$ constraint: $a+b=450$ $2\text{nd}$ constraint: $ab=M$ $b=450-a$, $M\left( a \right)=a\left( 450-a \right)=450a-{{a}^{2}}$ $\begin{align} &nbsp;&nbsp;& M\left( a \right)=450a-{{a}^{2}} \\ &nbsp;& \Rightarrow M'(a)=450-2a \\ &nbsp;& \Rightarrow 0=450-2a \\ &nbsp;& \Rightarrow a=225 \\ &nbsp;& \Rightarrow M''(a)=-2 \\ \end{align}$ $\begin{align} &nbsp;&nbsp;& b=450-225 \\ &nbsp;& b=225 \\ \end{align}$ $\begin{align} &nbsp;&nbsp;& ab=225\times 225 \\ &nbsp;& ab=50625 \\ \end{align}$

 If there are two numbers $f$ and $g$ such that their sum $2f+3g=60$. Given that $\left( 2f+1 \right)\left( 3g+2 \right)$ is a minimum. 

$2f+3g=60$ $M=\left( 2f+1 \right)\left( 3g+2 \right)$ $f=30-\dfrac{3}{2}g$, $M=\left( 2\left( 30-\dfrac{3}{2}g \right)+1 \right)\left( 3g+2 \right)=\left( 59-3g \right)\left( 3g+2 \right)$ $M=-9{{g}^{2}}+171g+118$ $\begin{align} &nbsp;&nbsp;& M'=-18g+171=0 \\ &nbsp;& \Rightarrow g=9.5 \\ \end{align}$ $M''=-18$ $\begin{align} &nbsp;&nbsp;& f=30-\dfrac{3}{2}\times \dfrac{19}{2} \\ &nbsp;& \Rightarrow f=\dfrac{120}{4}-\dfrac{57}{4} \\ &nbsp;& \Rightarrow f=\dfrac{63}{4},g=\dfrac{19}{2} \\ \end{align}$

 Three numbers, $l,w,h$ are being used If two of these numbers are equal and one of them is not equal to the other two. Calculate the values of each variable, $l$, $w$ and $h$ of such that the product of these is a maximum and $l=w$.

 Using the information given in the question, set up $2$ constraints and express each in terms of $2$ variables. Calculate the $1\text{st}$ and $2\text{nd}$ derivative of the function that has a maximum and solve for each variable. Since two of the variables are the same, one of the variables can be expressed as the other.

$\begin{align} &nbsp;&nbsp;& 60=lw+2\left( lh \right)+2\left( wh \right) \\ &nbsp;& 60=2wh+{{w}^{2}}+2wh \\ &nbsp;& 60={{w}^{2}}+4wh \\ \end{align}$ $\begin{align} &nbsp;&nbsp;& V=lwh \\ &nbsp;& l=w \\ &nbsp;& V={{w}^{2}}h \\ \end{align}$ $\begin{align} &nbsp;&nbsp;& 60-{{w}^{2}}=4wh \\ &nbsp;& h=\dfrac{60-{{w}^{2}}}{4w} \\ \end{align}$ $\begin{align} &nbsp;&nbsp;& V=\left( \dfrac{60-{{w}^{2}}}{4w} \right){{w}^{2}} \\ &nbsp;& V=\dfrac{1}{4}\left( 60-{{w}^{2}} \right)w=\dfrac{1}{4}\left( -{{w}^{3}}+60w \right) \\ \end{align}$ $\begin{align} &nbsp;&nbsp;& V'=-\dfrac{3}{4}{{w}^{2}}+60=0 \\ &nbsp;& w=4\sqrt{5} \\ &nbsp;& V''=-\dfrac{3}{2}w \\ \end{align}$ $\begin{align} &nbsp;&nbsp;& l=w=4\sqrt{5} \\ &nbsp;& h=\dfrac{60-{{\left( 4\sqrt{5} \right)}^{2}}}{4\left( 4\sqrt{5} \right)} \\ &nbsp;& h=-\dfrac{\sqrt{5}}{4} \\ \end{align}$