I hate these I don't understand how to do them even the teacher said they are hard what r some tips to solve these
A rectangle has it's base on x axis and upper to verticies on the parabola y=12-x^2
Wats the max area the rectangle can have
I know what the curve looks like and I know I need to use the area equation but that's all I know thanks ten points
A rectangle has it's base on x axis and upper to verticies on the parabola y=12-x^2
Wats the max area the rectangle can have
I know what the curve looks like and I know I need to use the area equation but that's all I know thanks ten points
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Suppose you put one vertex at the point (x, 0) (note that this point is on the x-axis). Since y = 12 - x^2 is an even function, we want to put the other vertex that is on the x-axis at (-x, 0). The other two vertices are at the points (-x, 12 - x^2) and (x, 12 - x^2).
The base of the rectangle is the distance between (-x, 0) and (x, 0), which is 2x, and the height of the rectangle is 12 - x^2. So, the area of the rectangle is:
A = 2x(12 - x^2).
From here, you can maximize A using derivatives.
I hope this helps!
The base of the rectangle is the distance between (-x, 0) and (x, 0), which is 2x, and the height of the rectangle is 12 - x^2. So, the area of the rectangle is:
A = 2x(12 - x^2).
From here, you can maximize A using derivatives.
I hope this helps!
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y =12 - x^2 ........(i)
A=Area = xy ...........(ii)
from (i)
A =x(12-x^2)=12x-x^3
Differentiate
dA/dx = 12-3x^2
For Max/Minimum dA/dx =0
12-3x^2 =0 or x=2
Differentiate again
d^2A/dx^2 =-6x
Put x = 2 then d^a/dx^2 = -12 <0
Therefore at x =2 the Area is Maximum
Therefore Max Area = xy =2(12-4) =2(8) =16 unit^2.........Ans
A=Area = xy ...........(ii)
from (i)
A =x(12-x^2)=12x-x^3
Differentiate
dA/dx = 12-3x^2
For Max/Minimum dA/dx =0
12-3x^2 =0 or x=2
Differentiate again
d^2A/dx^2 =-6x
Put x = 2 then d^a/dx^2 = -12 <0
Therefore at x =2 the Area is Maximum
Therefore Max Area = xy =2(12-4) =2(8) =16 unit^2.........Ans