There is 70m of fencing wire which is used to create 3 sides, brick wall is used for the 4th side. Calculate the maximum area of the rectangle, and state it's dimensions.
(I got the correct maximum area but can't figure out how to get the side lengths.)
I started with the perimeter.
P = 2x + y
70 = 2x + y
y = 70 - 2x
Then I got the area.
A = lw
A = (70 - 2x)(x)
From that I got max area or 612.5. And x = 0 and x = 35. If I substitute them in the initial equation (70 - 2x) or (x). I get one of them as 0. Please help!
As Steve pointed out to you in your previous posting of the same question, x = 0 and x = 35 are the values which give you an area of zero, which is not what you want.
A = (70-2x)(x) is represented by a parabola which opens downwards, so its vertex shows its maximum and at what x that maximum happens.
Steve pointed out that the x is halfway between 0 and 35
so x = 17.5 m
and 2x would be 35 m
That would give you an area of 312.5, which you got, and sides of 17.5 m and 35 m
check: 2(17.5) + 35 = 70
area = (70-35)(17.5) = 612.5
btw, you should have started the solution by defining what x and y represent.
I don't know how to reply directly to you. And I am sorry for posting this twice, I got an error the first time, so I posted it again but somehow it still posted it.
Thanks for your response. I have an exam tomorrow so wish me luck :D
To find the dimensions that will maximize the area of the rectangle, we can use calculus. Let's begin the process:
1. We have the equation for the area: A = (70 - 2x) * x.
2. To find the maximum area, we need to find the critical point where the derivative of the area equation equals zero.
3. Differentiate the area equation with respect to x:
dA/dx = 70 - 4x.
4. Now set dA/dx equal to zero and solve for x:
70 - 4x = 0.
4x = 70.
x = 70/4.
x = 17.5.
5. We have the x-coordinate for the critical point, which is x = 17.5.
6. To find the corresponding y-coordinate, substitute x = 17.5 back into the equation y = 70 - 2x:
y = 70 - 2(17.5).
y = 70 - 35.
y = 35.
7. Therefore, the critical point is (17.5, 35), representing the dimensions of the rectangle that maximize its area.
8. Substitute these dimensions back into the area equation to find the maximum area:
A = (70 - 2(17.5))(17.5).
A = (70 - 35)(17.5).
A = 35 * 17.5.
A = 612.5 square units.
So, the maximum area of the rectangle is 612.5 square units, with dimensions of 17.5 units for one side and 35 units for the other side.