You answer will be X=0 hope this helps
If A=38x-x^2 then
dA/dx=38-2x
d2A/dx2=-2
Since the acceleration, d2A/dx2 is a constant negative, when velocity, dA/dx=0, it will be an absolute maximum for A(x)
dA/dx=0 only when 38=2x, x=19
A(19)=38(19)-19^2
A(19)=722-361
A(19)=361 ft^2
So the maximum possible area is 361 ft^2
(This will always be true as the maximum possible area enclosed by a given amount of material will always be a perfect square...)
Answer:
The desired equation is y = (-8/3)x + 26/3.
Step-by-step explanation:
Moving from (1,6) to (4, -2) involves an increase of 3 in x and a decrease of 8 in y. Thus, the slope of the line thru these two points is m = rise / run = -8/3.
Using the slope-intercept form of the eq'n of a straight line and inserting the data given (slope = m = -8/3, x = 4, y = -2), we get:
y = mx + b => -2 = (-8/3)(4) + b, or -2 = -32/3 + b
Multiply all terms by 3 to clear out the fraction:
-6 = -32 + 3b.
Then 26 = 3b, and b = 26/3.
The desired equation is y = (-8/3)x + 26/3.
