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2 years ago

The perimeter of a rectangle is 24 inches. Find the dimension If the length of the rectangle is 3 inches more than it’s width.

Mathematics

1 answer:

GuDViN [60]2 years ago

8 0

Answer:

length is 7.5 and width is 4.5

Step-by-step explanation:

Its given that perimeter of the triangle is 24 units.

The condition is that length is 3 units more than the width.

l=w+3;

Formula for perimeter is 2(l+w).

so ,

2(w+w+3)=24,

2w+3=12,

w=4.5,

l=w+3=7.5

Hence for the given conditions length is 7.5 and width is 4.5.

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Answer:

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Step-by-step explanation:

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2 years ago

Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to f

tino4ka555 [31]

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :- f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x) to have a maximum or minimum at given point.

ii) find first derivative $f^{l} (x)$ and equating zero

iii) solve and find 'x' values

iv) Find second derivative $f^{ll}(x) >0$ then find the minimum value at x=a

v) Find second derivative $f^{ll}(x)$ then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

step1:- find first derivative $f^{l} (x)$ and equating zero

$f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)$

$f^{l}(x) = \frac{1}{x^2+1} (2x)$ ……………(1)

$f^{l}(x) = \frac{1}{x^2+1} (2x)=0$

the point is x=0

step2:-

Again differentiating with respective to 'x', we get

$f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}$

on simplification , we get

$f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}$

put x= 0 we get $f^{ll}(0) = \frac{2}{(1)^2}$ > 0

$f^{ll}(x) >0$ then find the minimum value at x=0

Final answer:-

The minimum value of the given function is f(0) = 0

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