A walkway is 4 feet wide and is paved with square stones that are eight inches wide

1. A walkway is 4 feet wide and is paved with square stones that are eight inches wide. How many stones wide is the path?

  • a. 3
  • b. 10
  • c. 6
  • d. 12
  • e. 4

2. At which points on the curve y = 1 + 40x3 – 3x5 does the tangent line have the largest slope?

(x,y) Smaller x-value

(x,y) Larger x-value

The correct answer and explanation is:

Let’s solve each problem step by step.

Problem 1: Walkway Width in Stones

  • The walkway is 4 feet wide.
  • The stones are 8 inches wide.
  • Convert feet to inches: 4 feet=4×12=48 inches4 \text{ feet} = 4 \times 12 = 48 \text{ inches}
  • Now, determine how many stones fit into the width: 48 inches8 inches per stone=6\frac{48 \text{ inches}}{8 \text{ inches per stone}} = 6
  • The correct answer is (c) 6.

Problem 2: Finding Points with Maximum Slope

Given the function: y=1+40×3−3x5y = 1 + 40x^3 – 3x^5

The slope of the tangent line is given by the derivative: dydx=120×2−15×4\frac{dy}{dx} = 120x^2 – 15x^4

To find the points where the tangent has the largest slope, we differentiate again to find critical points of dy/dxdy/dx: d2ydx2=240x−60×3\frac{d^2y}{dx^2} = 240x – 60x^3

Set the second derivative equal to zero: 240x−60×3=0240x – 60x^3 = 0

Factor: 60x(4−x2)=060x(4 – x^2) = 0

Solve for xx: x=0orx=±2x = 0 \quad \text{or} \quad x = \pm2

To determine where the slope is maximum, we check d2y/dx2d^2y/dx^2 at these points. The largest positive value of d2y/dx2d^2y/dx^2 occurs at x=−2x = -2 and x=2x = 2, indicating maximum slopes.

Find the corresponding y-values: y(−2)=1+40(−2)3−3(−2)5=1+40(−8)−3(−32)y(-2) = 1 + 40(-2)^3 – 3(-2)^5 = 1 + 40(-8) – 3(-32) =1−320+96=−223= 1 – 320 + 96 = -223 y(2)=1+40(2)3−3(2)5=1+40(8)−3(32)y(2) = 1 + 40(2)^3 – 3(2)^5 = 1 + 40(8) – 3(32) =1+320−96=225= 1 + 320 – 96 = 225

Thus, the points where the tangent has the largest slope are:

  • Smaller x-value: (-2, -223)
  • Larger x-value: (2, 225)

Now, I will generate an image of the function y=1+40×3−3x5y = 1 + 40x^3 – 3x^5 showing the points where the maximum slope occurs.

Here is the graph of the function y=1+40×3−3x5y = 1 + 40x^3 – 3x^5, highlighting the points (−2,−223)(-2, -223) and (2,225)(2, 225) where the tangent line has the largest slope. Let me know if you need any further explanation!

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