Carmen is at a used bookstore. Paperback books cost $ 1 each, and hardcover books cost $ 2 each. Carmen can spend up to $ 10 in all.A. Write an inequality that represents this situation. let x be the number of paperback books, and let Y be the number of hardcover books.
B. Graph the solution to your inequality on the plane below

Answer:

a) Each corral should be 33⅓ ft long and 25 ft wide

b) The total enclosed area is 1666⅔ ft²

Step-by-step explanation:

I assume that the corrals have identical dimensions and are to be fenced as in the diagram below

Let x = one dimension of a corral

and y = the other dimension

(a) Dimensions to maximize the area

The total length of fencing used is:

4x + 3y = 200

4x = 200 – 3y

x = 50 - ¾y

The area of one corral is A = xy, so the area of the two corrals is

A = 2xy

Substitute the value of x

A = 2(50 - ¾y)y

A = 100 y – (³/₂)y²

This is the equation for a downward-pointing parabola:

A = (-³/₂)y² + 100y

a = -³/₂; b = 100; c = 0

The vertex (maximum) occurs at  

y = -b/(2a)  = 100 ÷ (2׳/₂) = 100 ÷ 3 = 33⅓ ft  

4x + 3y = 100

Substitute the value of y

4x + 3(33⅓) = 200

4x + 100 = 200

4x = 100  

x = 25 ft

Each corral should measure 33⅓ ft long and 25 ft wide.

Step 2. Calculate the total enclosed area

The enclosed area is 50 ft long and 33⅓ ft wide.

A = lw = 50 × 100/3 = 5000/3 = 1666⅔ ft²

Final answer:

The maximum area is achieved when the shared fence is 50 feet and the other two sides are 75 feet each, yielding a maximum area of 3750 square feet.

Explanation:

This problem can be solved by the principles of calculus. Assuming that the two corrals share a common side, we can say the total length of fencing is divided into two lengths (x and y). The optimization problem can be formed as follows:

• x = length of the common fence
• y = length of the other sides

Since the total length available is 200 feet, 2y + x = 200. The area A = xy. Substitute y=(200-x)/2 into the area formula to get a quadratic A = x(200-x)/2. This graph opens downwards, meaning the vertex is the maximum point. The x-coordinate of the vertex of a quadratic given in standard form like Ax^2 + Bx + C is -B/2A. Therefore, x = -B/2A = 200/(2*2) = 50. Substitute x back into y = (200-2x)/2 to get y = 75. So, the maximum area is achieved with a common side of 50 feet and the other sides being 75 feet each.

The maximum area A can be found by substituying these values back into the area formula: A = 75*50 = 3750 square feet.

Learn more about Optimization here:

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Answer:

B. 88!!!

Step-by-step explanation:

Answer:

None

Step-by-step explanation:

I said this because i usually don't find any solutions for graphs until i find out that there are infinite or one

Answer: 48 Boxes

Step-by-step explanation:

20% of 480 = 0.20 x 480 = 96 cracked  

480 - 96 = 384

384 ÷ 8 = 48 boxes