Answer:
Dimensions of page should be width of 6 inches and height of 9 inches
Step-by-step explanation:
Let x be the width of the printed part in inches
Let y be height of the printed part in inches.
Thus, Area of printed part; A = xy
And area of printed part is given as 24.
Thus, xy = 24
Making y the subject, we have;
y = 24/x
Now, the question says the top and bottom margins are 1.5 inches.
Thus, width of page = x + 1 + 1 = x + 2
And also the margins on each side are both 1m in length, thus the height of page will be:
y + 1.5 + 1.5 = y + 3
So area of page will now be;
A = (x + 2)•(y+3)
From earlier, we got y = 24/x
Thus,plugging this into area of page, we have;
A = (x + 2)•((24/x)+3)
A = 24 + 3x + 48/x + 6
A = 30 + 3x + 48/x
For us to find the minimum dimensions, we have to find the derivative of A and equate to zero
Thus,
dA/dx = 3 - 48/x²
Thus, dA/dx = 0 will be
3 - 48/x² = 0
Multiply through by x²:
3x² - 48 = 0
Thus,
3x² = 48
x² = 48/3
x = √16
x = 4 inches
Plugging this into y = 24/x,we have;
y = 24/4 = 6 inches
We want dimensions of page at x = 4 and y = 6.
From earlier, width of page = x + 2.
Thus,width = 4 + 2 = 6 inches
Height = y + 3 = 6 + 3 = 9 inches
So dimensions of page should be width of 6 inches and height of 9 inches
= 75 minutes. Now we divide 40 by 75. 40÷75=0.53333333333333 miles. We know 1 hour is 60 minutes so we are going to multiply 0.5333333333333 by 60. 60×0.533333333333=32 miles. Therefore, your answer is 32 miles.