Sum of two positive numbers are 20. Find those numbers if the sum of one number squared and other number cubed is lowest possible.
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Let x be one number. The other number is 20-x
So you want to minimize (20-x)^2 + x^3
f(x) = x^3 + (20-x)^2
f'(x) = 3x^2 -2(20-x)=0
3x^2 +2x - 40=0
(x+4)(3x-10)=0
x= -4 or x= 10/3
x must be positive so x= 10/3
The two numbers are 10/3 and 50/3
So you want to minimize (20-x)^2 + x^3
f(x) = x^3 + (20-x)^2
f'(x) = 3x^2 -2(20-x)=0
3x^2 +2x - 40=0
(x+4)(3x-10)=0
x= -4 or x= 10/3
x must be positive so x= 10/3
The two numbers are 10/3 and 50/3
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Hi,
we only need to find one number (x), the other one will automatically be 20-x
So we need to find the minimum of the function
f(x) = x² + (20-x)³ with x being in the interval (0; 20) because both numbers need to be positive.
In order to find the minimum, we need to find the zeros of the first derivative to see where the critical points are
f ' (x) = -3 x² +122 x-1200 = 0
-(-24 + x) (-50 + 3 x) = 0
x=24 or x= 50/3
24 is outside the given interval, so x=50/3
To check if that is really a minimum, we produce the second derivative
f '' (x) = -6x + 122
for x=50/3 the value of f '' is 22 which means that at x=50/3 we have a minimum
So the numbers are 50/3 and 20-50/3=60/3-50/3 = 10/3
we only need to find one number (x), the other one will automatically be 20-x
So we need to find the minimum of the function
f(x) = x² + (20-x)³ with x being in the interval (0; 20) because both numbers need to be positive.
In order to find the minimum, we need to find the zeros of the first derivative to see where the critical points are
f ' (x) = -3 x² +122 x-1200 = 0
-(-24 + x) (-50 + 3 x) = 0
x=24 or x= 50/3
24 is outside the given interval, so x=50/3
To check if that is really a minimum, we produce the second derivative
f '' (x) = -6x + 122
for x=50/3 the value of f '' is 22 which means that at x=50/3 we have a minimum
So the numbers are 50/3 and 20-50/3=60/3-50/3 = 10/3
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x + y = 20
Minimize x^3 + y^2
y = 20 - x
F(x) = x^3 + (20 - x)^2
Find critical numbers of F(x):
F'(x) = 3x^2 + 2(20 - x)(-1)
= 3x^2 - 40 + 2x
= (x + 4)(3x - 10)
F'(x) = 0 or F'(x) DNE
No DNE critical numbers
x = {-4, 10/3}
x>0, therefore, x = 10/3 and y = 50/3.
Check using 1st or 2nd derivative test.
Minimize x^3 + y^2
y = 20 - x
F(x) = x^3 + (20 - x)^2
Find critical numbers of F(x):
F'(x) = 3x^2 + 2(20 - x)(-1)
= 3x^2 - 40 + 2x
= (x + 4)(3x - 10)
F'(x) = 0 or F'(x) DNE
No DNE critical numbers
x = {-4, 10/3}
x>0, therefore, x = 10/3 and y = 50/3.
Check using 1st or 2nd derivative test.