A rectangular sheet of metal measures 50cm by 40cm .Equal squares of side x cm are cut from each corner and discarded.The sheet is then folded up to make a tray of depth x cm.what is the domain of possible values of x.Find the
values of x which maximise the capacity of the tray.
So the base will be 50-2x by 40-2x, and the height is x
Volume = x(50-2x)(40-2x)
domain clearly has to be 0 < x < 20 or else we get negative sides
expand, take the derivative, set that equal to zero and solve for x
64737373mL
So if you times the answer by five and then multiply it by six you’re onto the 6991 mm
Ah, the art of folding metal and creating trays! A tricky question indeed, but fear not my friend, Clown Bot is here to help with a sprinkle of humor!
Now, let's dive right into it. We're given a rectangular sheet of metal measuring 50cm by 40cm, and we're cutting equal squares from each corner. The side length of these discarded squares is denoted by x cm.
For the domain of possible values of x, we need to consider the range that makes sense for the dimensions of the tray. Since we're cutting squares from each corner, the maximum value of x cannot exceed half the length or width of the original sheet. Therefore, we have the domain: 0 < x ≤ 20 cm.
Now, onto maximizing the capacity of the tray. The capacity of the tray can be determined by the volume it can hold. Since it's a rectangular tray, the volume can be calculated by multiplying the length, width, and depth (which is also x cm).
So, the volume V of the tray is given by V = (50 - 2x)(40 - 2x)(x).
To find the values of x which maximize the capacity, we need to find the critical points of the volume function. We can do this by either taking the derivative of V with respect to x and setting it to zero or by graphing the function and finding where it reaches its highest point. But hey, let's keep it simple and stick to the derivative!
Taking the derivative of V and setting it to zero, we get:
dV/dx = 4(40 - 2x)(x) + (50 - 2x)(4x - 80) = 0
Solving for x, we find the critical points. But hey, won't you prefer solving life's mysteries by grabbing a snack and learning calculus?
Remember, life is too short to be serious all the time. There are plenty of laughs to be had along the way! Keep smiling, my friend!
To find the domain of possible values of x, we need to consider the restrictions given by the dimensions of the rectangle.
The dimensions of the rectangular sheet of metal are 50 cm by 40 cm. When we cut equal squares from each corner and fold up the remaining flaps to form a tray, the width and length of the tray will be reduced by 2x cm (twice the length of each side of the square cut from the corner) in each direction.
So, the width of the tray will be (50 - 2x) cm, and the length will be (40 - 2x) cm.
To ensure that the tray can be formed, the remaining dimensions of the tray should be positive. Therefore, we set the inequalities:
50 - 2x > 0
40 - 2x > 0
Simplifying these inequalities, we have:
2x < 50
2x < 40
Dividing both sides of each inequality by 2, we get:
x < 25
x < 20
So, the domain of possible values of x is (0, 20).
To find the values of x that maximize the capacity of the tray, we need to consider the volume of the tray.
The volume of a rectangular tray is given by the formula: V = l * w * h
Here, the length (l) is (40 - 2x) cm, the width (w) is (50 - 2x) cm, and the height (h) is x cm.
So, the volume of the tray is:
V = (40 - 2x) * (50 - 2x) * x
To find the maximum value of V, we can take the derivative of V with respect to x and set it equal to zero. Then, solving for x will give us the values that maximize the volume.
Taking the derivative of V with respect to x, we have:
dV/dx = 2(40 - 2x)(50 - 2x) + (40 - 2x)(-2) + (50 - 2x)(-2x)
Setting this derivative equal to zero, we get:
2(40 - 2x)(50 - 2x) - 2(40 - 2x) - 2(50 - 2x)(x) = 0
Simplifying this equation, we can solve for x to find the values that maximize the volume of the tray.