Complete question is;
A model for a company's revenue from selling a software package is R = -2.5p² + 500p, where p is the price in dollars of the software. What price will maximize revenue? Find the maximum revenue.
Answer:
Price to maximize revenue = $100
Maximum revenue = $25000
Step-by-step explanation:
We are told that:
R = -2.5p² + 500p, where p is the price in dollars of the software.
The maximum revenue will occur at the vertex of the parabola.
Thus, the price at this vertex is;
p = -b/2a
Where a = - 2.5 and b = 500
Thus:
p = -500/(2 × -2.5)
p = -500/-5
p = 100 in dollars
Maximum revenue at this price is;
R(100) = -2.5(100)² + 500(100)
R(100) = -25000 + 50000
R(100) = $25000
Answer:
-4 and 2
Step-by-step explanation:
The exact numbers of that position is -4 and 2
It takes 8 liters to cover 800 sqft of floor
there are 3.78541 liters in a gallon so it would take 2.11338 gallons to cover the floor
![\bf -\cfrac{1}{4}(y+2)^2=(x-1)\implies (y+2)^2={-4}(x-1) \\\\\\\ [y-(-2)]^2=-4(x-1)\quad \begin{cases} k=-2\\ h=1\\ 4p=-4 \end{cases}\implies 4p=-4\implies \boxed{p=-1}](https://tex.z-dn.net/?f=%5Cbf%20-%5Ccfrac%7B1%7D%7B4%7D%28y%2B2%29%5E2%3D%28x-1%29%5Cimplies%20%28y%2B2%29%5E2%3D%7B-4%7D%28x-1%29%0A%5C%5C%5C%5C%5C%5C%5C%0A%5By-%28-2%29%5D%5E2%3D-4%28x-1%29%5Cquad%20%0A%5Cbegin%7Bcases%7D%0Ak%3D-2%5C%5C%0Ah%3D1%5C%5C%0A4p%3D-4%0A%5Cend%7Bcases%7D%5Cimplies%204p%3D-4%5Cimplies%20%5Cboxed%7Bp%3D-1%7D)
so, is a horizontal parabola, the "p" distance is 1, however, we ended up with a negative value, that means, the parabola is opening to the left-hand-side, with a vertex at (1, -2), and its focus at 0, -2, like you see in the picture below, one unit to the left of the vertex.
