To answer this item, we take the differential of the equation and equate to zero.
C(x) = 0.8x² - 256x + 25939
Differentiation,
dC(x) = 1.6x - 256
dC(x) = 1.6x - 256 = 0
The value of x from the derived equation above is 160.
Thus, the number of machine to be made in order to minimize the cost should be 160.
Answer:
y = (3/4)x + 2
Step-by-step explanation:
Slope-intercept form is y=mx+b where (x, y) is a point on the linear graph, m is the slope (rise/run), and b is the y-intercept (the y-value at which the graph passes through the y-axis).
Looking at the graph, we can see that the point at which the line crosses the y-axis is (0, 2) which makes it the y-intercept. Thus, the b in the slope-intercept form is 2.
Next, we are looking for the slope of the line. To do this, we can calculate the rise/run of the line by choosing to points on it. Since we already have the point (0, 2), we just need one more.
For example, the point (-4, -1) can be used. The slope can be found by ((y-y)/(x-x)) in which the first y and x values correspond with the first point and that of the second correspond with the second set. So in this case, m = (2-(-1))/(0-(-4)) = 3/4
Plugging in the calculated m and b value in the slope intercept equation, we get y = (3/4)x + 2
Sec^2 x - 1 = tan^2x
Proof:
Sec^2x = 1+ tan^2x
1/cos^2x = 1 + sin^2x/cos^2x
<span>1/cos^2x - sin^2x/cos^2x = 1
</span>Using common denominator:
(1-sin^2x)/cos^2x = 1
sin^2x + cos^2 x = 1
cos^2 x = 1 - sin^2x
Substituting :
cos^2x/<span>cos^2x = 1
</span>1 = 1
Left hand side = right hand side
Area=length*width
let the sides perpendicular to the river be x
then the side parallel to the river is 4,700-2x
A(x)=x(4700-2x)
A(x)=4700x-2x^2
This a quadratic function with a=-2 and b=4700
thus
Maximum area occurs where x=-b/2a=-4700(-2*2)=1,175 ft
length=4700-2(1175)=2,350 ft
