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3 years ago

What is equivalent ratio of 5:8:4 and explain how u got it

Mathematics

1 answer:

igor_vitrenko [27]3 years ago

6 0

this is example

So, first we need to write the given ratio as fraction,

= 8/18

= (8 × 2)/(18 × 2)

= 16/36

= 16 : 36 (one equivalent ratio),

So, 16 : 36 is an equivalent ratio of 8 : 18.

Now we will find another equivalent ratio of 8 : 18 by using division.

Similarly, first we need to write the given ratio as fraction,

= 8/18

= (8 ÷ 2)/(18 ÷ 2)

= 4/9

= 4 : 9 (another equivalent ratio)

So, 4 : 9 is an equivalent ratio of 8 : 18.

Therefore, the two equivalent ratios of 8 : 18 are 16 : 36 and 4 : 9.

2. Frame two equivalent ratios of 4 : 5.

Solution:

To find two equivalent ratios of 4 : 5 we need to apply multiplication method only to get the answer in integer form.

First we need to write the given ratio as fraction,

= 4/5

= (4 × 2)/(5 × 2)

= 8/10

= 8 : 10 is one equivalent ratio,

Similarly again, we need to write the given ratio 4 : 5 as fraction to get another equivalent ratio;

= 4/5

= (4 × 3)/(5 × 3)

= 12/15 is another equivalent ratio

Therefore, the two equivalent ratios of 4 : 5 are 8 : 10 and 12 : 15.

Note: In this question we can’t apply division method to get the answer in integer form because the G.C.F. of 4 and 5 is 1. That means, 4 and 5 cannot be divisible by any other number except 1.

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$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).