# when to use dy/dx=0Watch

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#1
the question is "A piece of wire is bent to form a circle with radius 7cm. If the same wire is used to form a rectangle, find the maximum area of the rectangle."

length of wire=2(22/7)(7)=44
then let length of rec=x and width=y
2x+2y=44
y=22-x
so the area of rec=xy
i then simultaneous both equations
so i was left with A=22x-x^2
then i differentiate it and i got dA/dx=22-2x
then i was told to let dA/dx=0, then find the value of x and y, and sub it back to the length and width, then multiply to get 121, which is the answer.

But i still dont understand why do i need to use dA/dx=0
0
4 weeks ago
#2
(Original post by khorrr)
the question is "A piece of wire is bent to form a circle with radius 7cm. If the same wire is used to form a rectangle, find the maximum area of the rectangle."

length of wire=2(22/7)(7)=44
then let length of rec=x and width=y
2x+2y=44
y=22-x
so the area of rec=xy
i then simultaneous both equations
so i was left with A=22x-x^2
then i differentiate it and i got dA/dx=22-2x
then i was told to let dA/dx=0, then find the value of x and y, and sub it back to the length and width, then multiply to get 121, which is the answer.

But i still dont understand why do i need to use dA/dx=0
Your aim is to find the maximum value that A = 22x - x^2 can take as x varies. If you consider the graph then this is the maximum point of the graph. To find the maximum you can either complete the square of set the first derivative equal to 0.

Have you found maximum/minimums using differentiation before? The explanation will be in your textbook.
0
#3
(Original post by Notnek)
Your aim is to find the maximum value that A = 22x - x^2 can take as x varies. If you consider the graph then this is the maximum point of the graph. To find the maximum you can either complete the square of set the first derivative equal to 0.

Have you found maximum/minimums using differentiation before? The explanation will be in your textbook.
I've used dy/dx=0 to find the maximum/minimum point before. But for this question there is no graph involved, that's why i'm confused now why do i have to use dA/dx=0 to find the value of x...
0
4 weeks ago
#4
(Original post by khorrr)
I've used dy/dx=0 to find the maximum/minimum point before. But for this question there is no graph involved, that's why i'm confused now why do i have to use dA/dx=0 to find the value of x...
Take A as y.
A=22x-x^2
y=22x - x^2
y=x(22-x)

Thats a graph that you'd get, so using that length of wire you can make a rectangle with area.
So, the area of the rectangle can be anywhere from 0 to 121 cm^2
You can also see that the area can be 0 two times depending on the x value.

So, when the length of x is 0, the area is 0. (logical)
when the length of x is 22 (the entire length of wire) then the area is 0. (logical again)

There is still a graph there, but it's in the background, this is the same for all areas made with a finite length of anything.
1
#5
(Original post by Relentas)
Take A as y.
A=22x-x^2
y=22x - x^2
y=x(22-x)

Thats a graph that you'd get, so using that length of wire you can make a rectangle with area.
So, the area of the rectangle can be anywhere from 0 to 121 cm^2
You can also see that the area can be 0 two times depending on the x value.

So, when the length of x is 0, the area is 0. (logical)
when the length of x is 22 (the entire length of wire) then the area is 0. (logical again)

There is still a graph there, but it's in the background, this is the same for all areas made with a finite length of anything.
My doubt is clear now, thanks for your detailed explanation and patience
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