R(p) = -15p² + 200p + 10,000
To find the maximum of the function, we take the first derivative and equate to 0:
R'(p) = -30p + 200
0 = -30p + 200
p = 6.67
A price of $6.67 will maximize revenue.
Jacob wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn,so he needs no fence on that side.
Let w be the width of the enclosure (perpendicular to the barn) and let l be the length of the enclosure (parallel to the barn).
one side of the length is not counted for perimeter because one side of length will be against the barn.
Perimeter = 400 ft
Perimeter of rectangle = L + W + W
400 = L + 2W
L = 400 - 2W
Area = L * W
Replace L by 400 - 2W
A(W) = (400 - 2W) * W
Now we find out x coordinate of vertex to find the width that maximize the area
a= -2 and b = 400
The width w would maximize the area is w = 100ft
To find maximum area we plug in 100 for W in A(W)
the maximum area is 20,000 square feet
Answer:
._. -_- -.- .-. --__--
Step-by-step explanation:
Answer:
10.5 cm^2
Step-by-step explanation:
Area of a triangle = 1/2(L x H)
Reading from the diagram
length = 7 cm
height = 3 cm
therefore
Area of the triangle = 1/2( 7×3)
Area = 10.5 cm^2
54 - 18n
Multiply -9 by -6 to get 54.
Multiply -9 by 2n to get 18n.