Respuesta :
hello :
xy = 675...(1)
x >0 and y >0
by (1) : y = 675/x
x+3y = x+3(675)/x = x +2025/x
let : f(x) = x +2025/x
calchlate the minumum:
f'(x) = 1- 2025/x²
f'(x) = (x²-2025)/x²
f'(x) = 0 : x²-2025 =0
x² =2025 =45²
x = 45.... ( x >0 )
sign of f'(x) : -∞ـــــــــــــــــــــــــــــــــــــــــ 45ــــــــــــ/0/ـــــــــــ 45 -ـــــــــــــــــــ>+ ∞
++++++++ 0 --------- ---------0++++++++++++++++++
f(45) is a minumum value
f(45) = 90 ( minimize the sum of the first and three times the second)
xy = 675...(1)
x >0 and y >0
by (1) : y = 675/x
x+3y = x+3(675)/x = x +2025/x
let : f(x) = x +2025/x
calchlate the minumum:
f'(x) = 1- 2025/x²
f'(x) = (x²-2025)/x²
f'(x) = 0 : x²-2025 =0
x² =2025 =45²
x = 45.... ( x >0 )
sign of f'(x) : -∞ـــــــــــــــــــــــــــــــــــــــــ 45ــــــــــــ/0/ـــــــــــ 45 -ـــــــــــــــــــ>+ ∞
++++++++ 0 --------- ---------0++++++++++++++++++
f(45) is a minumum value
f(45) = 90 ( minimize the sum of the first and three times the second)
We want to minimize the sum of two numbers given that we know its product, we will find that the minimum of the sum is 90
Finding the equations:
First, we need to define the variables we will be using, let's define A as the first number and B as the second number.
Then we can write the product as:
A*B = 675
And we want to minimize:
A + 3*B
Minimizing the sum:
To do it, we first need to isolate one of the two numbers in the product, let's isolate A:
A = 675/B
Now we can replace this on the sum to get:
675/B + 3*B
To minimize this we just need to find the zeros of the first derivate, the first derivate is:
-675/B^2 + 3
And we need to make this equal to zero:
-675/B^2 + 3 = 0
675/B^2 = 3
675/3 = B^2
225 = B^2
√225 = B = 15
So the value of B that minimizes the sum is B = 15, and the minimum sum is:
675/15 + 3*15 = 90
If you want to learn more about minimizing, you can read:
https://brainly.com/question/25207394