THE AMOUNT OF CLOTH USED TO MAKE FOUR CURTAINS IS GIVEN BY THE FUNCTION A=-4X^2 + 40X WHERE X IS THE WIDTH OF ONE CURTAIN IN FEET AND a IS THE TOTAL AREA IN SQUARE FEET. FIND THE WIDTH THAT MAXIMIZES THE AREA OF THE CURTAINS. WHAT IS THE MAXIMUM AREA?
this problem consists of finding the vertex of the parabola,
the x of the vertex is -b/(2a) = -40/-8 = 5
when x = 5
A = -4(25) + 40(5) = 100
So the maximum area is 100 ft^2 , when the width is 5 ft.
I would suggest you have your caps lock fixed on your keyboard, it will be so much easier to read your post.
A=-4x^2+40x
By completing the square
A= -4(x^2-10x+25)+100
A= -4(x-5)^2+100
Vertex is (5,100)
Max is 100 ft
I do not know
Ah, the age-old question of maximizing curtain area. Let's find the width that makes those curtains as fabulous as possible!
To find the width that maximizes the area, we need to find the vertex of the quadratic function A = -4X^2 + 40X. The vertex occurs at the value of X that makes the quadratic function reach its maximum point.
The X-coordinate of the vertex can be found using the formula X = -b/2a, where a = -4 and b = 40. Plugging in these values, we get X = -40/2(-4) = 40/8 = 5 feet.
Now that we know the width is 5 feet, we can substitute this value into the function to find the maximum area. A = -4(5)^2 + 40(5) = -4(25) + 200 = -100 + 200 = 100 square feet.
So, the width that maximizes the area of the curtains is 5 feet, and the maximum area is 100 square feet.
Now you can curtainly make an informed decision!
To find the width that maximizes the area of the curtains and the maximum area, we need to find the vertex of the quadratic function A = -4X^2 + 40X. The vertex of a quadratic function is the point at which it reaches its maximum or minimum value. In this case, we are interested in the maximum value.
The function A = -4X^2 + 40X represents a downward-opening parabola because the coefficient of the X^2 term (-4) is negative. The X-coordinate of the vertex (the width that maximizes the area) can be found using the formula X = -b / (2a), where a is the coefficient of X^2 (-4) and b is the coefficient of X (40).
X = -b / (2a) = -(40) / (2(-4)) = -40 / -8 = 5
So the width that maximizes the area of the curtains is 5 feet.
To find the maximum area, we substitute this width (X = 5) back into the function A = -4X^2 + 40X:
A = -4(5)^2 + 40(5) = -4(25) + 200 = -100 + 200 = 100
Therefore, the maximum area of the curtains is 100 square feet.