ok, firstly, i dont really want to know the answer, i want to know the process of how to obtain the answer. heres my problem.
"Suppose that F = 16x + y and xy = 25, where x > 0 and y >0 at what values of x and y is the quantity F minimized?
so i see he did some substitution...
then he found derivative?!? and then using trick set it to so there was an x in the numerator so that when he set the derivative equal to zero he could obtain the x's.. he picks positive because x >0
then he finds second derivative and says concave up..
at the end the answer ends up being y = 25/ ( 5/4) = 20
plz help me figure out how to do all this.. i do give best answers, thanks in advance
"Suppose that F = 16x + y and xy = 25, where x > 0 and y >0 at what values of x and y is the quantity F minimized?
so i see he did some substitution...
then he found derivative?!? and then using trick set it to so there was an x in the numerator so that when he set the derivative equal to zero he could obtain the x's.. he picks positive because x >0
then he finds second derivative and says concave up..
at the end the answer ends up being y = 25/ ( 5/4) = 20
plz help me figure out how to do all this.. i do give best answers, thanks in advance
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F = 16x + y
xy = 25
y = 25/x
F = 16x + 25/x
dF/dx = 16 - 25/x^2
d2F/dx^2 = 50/x^3
We take the first derivative to determine the slope of F for some point (x , y). Wherever dF/dx = 0, we'll have either a minimum value of a maximum value because the function is neither increasing or decreasing at that point
Find when dF/dx = 0
16 - 25/x^2 = 0
16 = 25/x^2
x^2 = 25/16
x = -5/4 , 5/4
We take the second derivative to determine the concavity of the function. If it's concave up at a critical point (where the first derivative is 0), then we're looking at a minimum value. If it's concave down at a critical point, then we're looking at a maximum value.
50/x^3
50 / (-5/4)^3 => 50 / (-125 / 64) = 50 * 64 / (-125) = -25 * 64 * 2 / (25 * 5) = -128/5
50 / (5/4)^3 = 128/5
When x = -5/4, d2f/dx^2 is negative, so the function is concave down, therefore we're looking at a maximum point.
When x = 5/4, d2f/dx^2 is positive, so the function is concave up, therefore we're looking at a minimum point
x = 5/4
xy = 25
(5/4) * y = 25
y = 25 * (4/5) = 5 * 4 = 20
So our point is at (5/4 , 20)
xy = 25
y = 25/x
F = 16x + 25/x
dF/dx = 16 - 25/x^2
d2F/dx^2 = 50/x^3
We take the first derivative to determine the slope of F for some point (x , y). Wherever dF/dx = 0, we'll have either a minimum value of a maximum value because the function is neither increasing or decreasing at that point
Find when dF/dx = 0
16 - 25/x^2 = 0
16 = 25/x^2
x^2 = 25/16
x = -5/4 , 5/4
We take the second derivative to determine the concavity of the function. If it's concave up at a critical point (where the first derivative is 0), then we're looking at a minimum value. If it's concave down at a critical point, then we're looking at a maximum value.
50/x^3
50 / (-5/4)^3 => 50 / (-125 / 64) = 50 * 64 / (-125) = -25 * 64 * 2 / (25 * 5) = -128/5
50 / (5/4)^3 = 128/5
When x = -5/4, d2f/dx^2 is negative, so the function is concave down, therefore we're looking at a maximum point.
When x = 5/4, d2f/dx^2 is positive, so the function is concave up, therefore we're looking at a minimum point
x = 5/4
xy = 25
(5/4) * y = 25
y = 25 * (4/5) = 5 * 4 = 20
So our point is at (5/4 , 20)
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at the point (x,y) where a function F is minimized its first derivative F'(x,y) = 0 and its second derivative is F''(x,y) > 0
Now F = 16x + 25/x, setting F' = 16 - 25/x^2 = 0 => x^2 = 25/16 => x = +5/4 or -5/4
Now use F'' > 0 => 25/x^3 > 0 => x>0
So x = 5/4 is the required answer => y = 25/x = 20
Now F = 16x + 25/x, setting F' = 16 - 25/x^2 = 0 => x^2 = 25/16 => x = +5/4 or -5/4
Now use F'' > 0 => 25/x^3 > 0 => x>0
So x = 5/4 is the required answer => y = 25/x = 20