Pls help!
dy/dx
Y=3x²-7x-4

And dy /dx
Y=7x
Pls
"Solve using first principle "

Reach me on my whatsapp 09018921071

ToksZee:
Pls help!
dy/dx
Y=3x²-7x-4

And dy /dx
Y=7x
Pls
"Solve using first principle "

Lim(as h tends To 0) (3(x+h)/\2-7(x+h)-4-3(x)/\2-7(x)-4)/h....simplify and Make Sure You Get Rid Of the h(denomator). Before Plugging 0. Do Same For The Second Question. Na java phone I De Use, i For Solve Am Upload.

ToksZee:
Pls help!
dy/dx
Y=3x²-7x-4

Δy = 3(x + Δx)² – 7(x + Δx) – 4 – 3x² + 7x + 4
= 6x(Δx) + 3(Δx)² – 7Δx

Δy/Δx = 6x + 3(Δx) – 7

The limit of 6x + 3Δx – 7 as Δx approaches 0 is 6x – 7. Hence dy/dx = 6x – 7.

And dy /dx
Y=7x
Pls
"Solve using first principle"

Δy = 7(x + Δx) – 7x
= 7(Δx)

Δy/Δx = 7.

The limit of 7 as Δx approaches 0 is 7. Hence dy/dx = 7.

That guy above me, I throway cap for you. You are not just about ur redpill stuff. You look brilliant.

Martinez39s:
Δy = 3(x + Δx)² – 7(x + Δx) – 4 – 3x² + 7x + 4
= 6x(Δx) + 3(Δx)² – 7Δx

Δy/Δx = 6x + 3(Δx) – 7

The limit of 6x + 3Δx – 7 as Δx approaches 0 is 6x – 7. Hence dy/dx = 6x – 7.

Δy = 7(x + Δx) – 7x
= 7(Δx)

Δy/Δx = 7.

The limit of 7 as Δx approaches 0 is 7. Hence dy/dx = 7.

bro I take God beg you
help me with this

A utility company has a small power plant that can produce a kilowatt hours of electricity daily at a cost of 10 - (x/10^5 )cents each for 0 < x < 8 x 10^5. Consumers will use 10^5(10 - p/2) kilowatt hours of electricity daily at a price of p cents per kilowatt hour. What price should the utility charge to maximize its profit?

Lordsagna:


bro I take God beg you
help me with this

A utility company has a small power plant that can produce a kilowatt hours of electricity daily at a cost of 10 - x/10^5 cents each for 0 < x < 8 x 10^5. Consumers will use 10^5(10 - p/2) kilowatt hours of electricity daily at a price of p cents per kilowatt hour. What price should the utility charge to maximize its profit?

someone should help pls

Lordsagna:


bro I take God beg you
help me with this

A utility company has a small power plant that can produce a kilowatt hours of electricity daily at a cost of 10 - x/10^5 cents each for 0 < x < 8 x 10^5. Consumers will use 10^5(10 - p/2) kilowatt hours of electricity daily at a price of p cents per kilowatt hour. What price should the utility charge to maximize its profit?

I don't understand your question. Restate it again. It seems like an optimisation problem in differential calculus.

Also clarify if you meant 10 – (x/10⁵) instead of (10 – x)/10⁵. I will solve the problem for you.

Martinez39s:
I don't understand your question. Restate it again. It seems like an optimisation problem in differential calculus.

Also clarify if you meant 10 – (x/10⁵) instead of (10 – x)/10⁵. I will solve the problem for you.

yeah it's on optimization but I'm stuck.
10 -(x/10^5)

See Attached files for solutions.

Lordsagna:


yeah it's on optimization but I'm stuck.
10 -(x/10^5)

I have gotten that it's 10 – (x/10⁵). For one last time, check your question again.

Martinez39s:
I have gotten that it's 10 – (x/10⁵). For one last time, check your question again.

nothing is wrong with the question bro

3-b

Lordsagna:

nothing is wrong with the question bro

3-b

Solving Optimization Problems

1.Understand the problem. Clearly identify what quantity is to be maximized or minimized.(in this case,we are looking for the price that will maximize the profit)

2.Create equations relevant to the context of the problem, using the information given. (One of these should describe the quantity to be optimized. We'll call this the fundamental equation. i.e P(x)=R(x)-C(x)=> P(x)=xS.P(x)-C.P(x),,,,,,,,,,
s.p(selling price),c.p(cost price),P(profit).
critical point is at P'(x)=0)

2.If the fundamental equation defines the quantity to be optimized as a function of more than one variable, reduce it to a single variable function using substitutions derived from the other equations.
3.Identify the domain of this function, keeping in mind the context of the problem. which is (0<x<8*10^5)=>[0, 8*10^5]
4.Find the extreme values of this function on the determined domain.
5.then finally Identify the values of all relevant quantities of the problem.


my little contribution in helping you solving the problem

Hamzasaid:


Solving Optimization Problems

1.Understand the problem. Clearly identify what quantity is to be maximized or minimized.(in this case,we are looking for the price that will maximize the profit)

2.Create equations relevant to the context of the problem, using the information given. (One of these should describe the quantity to be optimized. We'll call this the fundamental equation. i.e P(x)=R(x)-C(x)=> P(x)=xS.P(x)-C.P(x),,,,,,,,,,
s.p(selling price),c.p(cost price),P(profit).
critical point is at P'(x)=0)

2.If the fundamental equation defines the quantity to be optimized as a function of more than one variable, reduce it to a single variable function using substitutions derived from the other equations.
3.Identify the domain of this function, keeping in mind the context of the problem. which is (0<x<8*10^5)=>[0, 8*10^5]
4.Find the extreme values of this function on the determined domain.
5.then finally Identify the values of all relevant quantities of the problem.


my little contribution in helping you solving the problem

I'm grateful sir.

Lordsagna:


I'm grateful sir.

I no be sir oohh
have u gotten the answer?

Hamzasaid:

I no be sir oohh have u gotten the answer?

. yeah..I gat a large value not sure if it's right.

Solution.
y=3x² -7x-4
Change in y=change in x:
(y+Δy)=3(x+Δx)² -7(x+Δx)-4
Subtract y from both sides:
y+Δy-y=3(x² +2xΔx+Δx²)-7(x+Δx)-4-y
Where y=3x² -3x-4
y+Δy-y=3x²+6xΔx+3Δx²-7x-7Δx-4-(3x² -7x-4)
Δy=3x² +6xΔx+3Δx² -7x-7Δx-4-3x²+7x+4
Δy=6xΔx-7Δx+3Δx²
Divide both sides by Δx:
Δy/Δx=6xΔx/Δx -7Δx/Δx +3Δx²/Δx
Δy/Δx=6x -7 +3Δx
Lim(Δy/Δx) =6x-7 +3(0) Δx--> 0
Lim(Δy/Δx) =6x -7 .....Q. E. D Δx-->0