Your answer would be 4.23 as the answer will be a long one, estimated.
Answer:
a. 1620-x^2
b. x=810
c. Maximum value revenue=$656,100
Step-by-step explanation:
(a) Total revenue from sale of x thousand candy bars
P(x)=162 - x/10
Price of a candy bar=p(x)/100 in dollars
1000 candy bars will be sold for
=1000×p(x)/100
=10*p(x)
x thousand candy bars will be
Revenue=price × quantity
=10p(x)*x
=10(162-x/10) * x
=10( 1620-x/10) * x
=1620-x * x
=1620x-x^2
R(x)=1620x-x^2
(b) Value of x that leads to maximum revenue
R(x)=1620x-x^2
R'(x)=1620-2x
If R'(x)=0
Then,
1620-2x=0
1620=2x
Divide both sides by 2
810=x
x=810
(C) find the maximum revenue
R(x)=1620x-x^2
R(810)=1620x-x^2
=1620(810)-810^2
=1,312,200-656,100
=$656,100
Answer:
Depth of the rain gutter is 8 inches
Step-by-step explanation:
Let’s assume ‘x’ is the depth of the rain gutter
Then the width of the rain gutter can be written as 16 - 2x
Cross sectional area
A = depth x width
Substitute values
A = x*(16 - 2x)
A = 16x – 2x^2
Now according to axis of symmetry for maximum area x = -b/2a
x = -16/2*(-2)
x = 4 inches depth of rain gutter, substitute the value of x to get
Width of rain gutter 16 – 2(4) = 8 inches
Area of the rain gutter for maximum water flow
A = 4 * 8
A = 32 square inch.
Answer: 2 meters.
Step-by-step explanation:
Let w = width of the cement path.
Dimensions of pool : Length = 15 meters , width = 9 meters
Area of pool = length x width = 15 x 9 = 135 square meters
Along width cement path, the length of region = 
width = 
Area of road with pool = 
Area of road = (Area of road with pool ) -(area of pool)
![\Rightarrow\ 112 =4w^2+48w+135- 135\\\\\Rightarrow\ 112= 4w^2+48w\\\\\Rightarrow\ 4 w^2+48w-112=0\\\\\Rightarrow\ w^2+12w-28=0\ \ \ [\text{Divide both sides by 4}]\\\\\Rightarrow\ w^2+14w-2w-28=0\\\\\Rightarrow\ w(w+14)-2(w+14)=0\\\\\Rightarrow\ (w+14)(w-2)=0\\\\\Rightarrow\ w=-14\ or \ w=2](https://tex.z-dn.net/?f=%5CRightarrow%5C%20112%20%3D4w%5E2%2B48w%2B135-%20135%5C%5C%5C%5C%5CRightarrow%5C%20112%3D%204w%5E2%2B48w%5C%5C%5C%5C%5CRightarrow%5C%204%20w%5E2%2B48w-112%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B12w-28%3D0%5C%20%5C%20%5C%20%5B%5Ctext%7BDivide%20both%20sides%20by%204%7D%5D%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B14w-2w-28%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%28w%2B14%29-2%28w%2B14%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%28w%2B14%29%28w-2%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%20w%3D-14%5C%20or%20%5C%20w%3D2)
width cannot be negative, so w=2 meters
Hence, the width of the road = 2 meters.
You can write it as:
-4≥x≤6
*The '≥' and '≤' just means that it is greater than or less than but it can still be equal to the number you are comparing it to.
Hope this helps! :)
