The question is incomplete. Here is the complete question.
Find the measurements (the lenght L and the width W) of an inscribed rectangle under the line y = -
x + 3 with the 1st quadrant of the x & y coordinate system such that the area is maximum. Also, find that maximum area. To get full credit, you must draw the picture of the problem and label the length and the width in terms of x and y.
Answer: L = 1; W = 9/4; A = 2.25;
Step-by-step explanation: The rectangle is under a straight line. Area of a rectangle is given by A = L*W. To determine the maximum area:
A = x.y
A = x(-
)
A = -
To maximize, we have to differentiate the equation:
= 
(-)
= -3x + 3
The critical point is:
= 0
-3x + 3 = 0
x = 1
Substituing:
y = -x + 3
y = -.1 + 3
y = 9/4
So, the measurements are x = L = 1 and y = W = 9/4
The maximum area is:
A = 1 . 9/4
A = 9/4
A = 2.25
Answer:
32
Step-by-step explanation:
No of desks along length of the rectangle = 7 as it is 7 rows wide.
No of desks along breadth of the rectangle = 7 as it is 5 rows long.
Total number of students is product of these two numbers.
So number of students = 35.
But three desks are empty.
Total students = 35- 3
= 32.
The answer is 81. since (-3)^2 is 9. Therefore, 9x9=81.
Answer:
ASK YOUR QUESTION
Middle School Mathematics 5+3 pts
Evaluate the expression 3.14(a2 + ab) when a = 3 and b = 4. (Input decimals only, such as 12.71, as the answer.)
Report 13.07.2017
Answer
3.14(a^2 + ab) when a = 3 and b = 4
then
3.14(3^2 + 3*4)
= 3.14(9+12)
= 3.14(21)
= 65.94
15
Step-by-step explanation:
c=PI*d
divide both sides by PI to get d by itself
d = c/PI
