Let’s explore more real-world applications of polynomials in different fields and practice solving them step-by-step.
1. Physics: Projectile Motion
A stone is thrown vertically upward with an initial velocity of 20 m/s20 \, \text{m/s}. The height h(t)h(t) of the stone at time tt seconds is given by:
h(t)=−5t2+20t+25.h(t) = -5t^2 + 20t + 25.
Problem 1: Maximum Height
Find the time at which the stone reaches its maximum height and calculate the height.
Solution:
For a quadratic function h(t)=−5t2+20t+25h(t) = -5t^2 + 20t + 25, the maximum height occurs at the vertex t=−b2at = -\frac{b}{2a}, where a=−5a = -5, b=20b = 20.
t=−202(−5)=2 seconds.t = -\frac{20}{2(-5)} = 2 \, \text{seconds}.
Substitute t=2t = 2 into h(t)h(t):
h(2)=−5(2)2+20(2)+25.h(2) = -5(2)^2 + 20(2) + 25. h(2)=−20+40+25=45 meters.h(2) = -20 + 40 + 25 = 45 \, \text{meters}.
Answer: Maximum height = 45 meters45 \, \text{meters} at t=2 secondst = 2 \, \text{seconds}.
Problem 2: Time When Stone Hits the Ground
Find the time tt when h(t)=0h(t) = 0.
Solution:
Set h(t)=0h(t) = 0:
−5t2+20t+25=0.-5t^2 + 20t + 25 = 0.
Divide through by −5-5:
t2−4t−5=0.t^2 – 4t – 5 = 0.
Factorize:
(t−5)(t+1)=0.(t – 5)(t + 1) = 0. t=5, t=−1.t = 5, \, t = -1.
Since time cannot be negative, t=5 secondst = 5 \, \text{seconds}.
Answer: The stone hits the ground after 5 seconds5 \, \text{seconds}.
2. Economics: Profit Optimization
A company’s profit P(x)P(x), in thousand dollars, is given by:
P(x)=−2×2+12x−10,P(x) = -2x^2 + 12x – 10,
where xx is the number of products sold in hundreds.
Problem 1: Maximum Profit
Find the maximum profit and the number of products to be sold to achieve it.
Solution:
The maximum value of P(x)P(x) occurs at x=−b2ax = -\frac{b}{2a}, where a=−2a = -2, b=12b = 12.
x=−122(−2)=3 (hundreds of products).x = -\frac{12}{2(-2)} = 3 \, \text{(hundreds of products)}.
Substitute x=3x = 3 into P(x)P(x):
P(3)=−2(3)2+12(3)−10.P(3) = -2(3)^2 + 12(3) – 10. P(3)=−18+36−10=8 (thousand dollars).P(3) = -18 + 36 – 10 = 8 \, \text{(thousand dollars)}.
Answer: Maximum profit = 8,000 dollars8,000 \, \text{dollars}, achieved by selling 300 products300 \, \text{products}.
3. Biology: Population Modeling
The population P(t)P(t) of a certain species in a region is modeled as:
P(t)=−t2+10t+40,P(t) = -t^2 + 10t + 40,
where tt is time in years.
Problem 1: Maximum Population
Find the year when the population reaches its maximum and the maximum population.
Solution:
The maximum value of P(t)P(t) occurs at t=−b2at = -\frac{b}{2a}, where a=−1a = -1, b=10b = 10.
t=−102(−1)=5 years.t = -\frac{10}{2(-1)} = 5 \, \text{years}.
Substitute t=5t = 5 into P(t)P(t):
P(5)=−(5)2+10(5)+40.P(5) = -(5)^2 + 10(5) + 40. P(5)=−25+50+40=65.P(5) = -25 + 50 + 40 = 65.
Answer: Maximum population = 6565, reached in 5 years5 \, \text{years}.
4. Geometry: Optimization Problems
A rectangular garden has an area of 200 m2200 \, \text{m}^2. Its length ll is 10 meters more than its width ww. Find the dimensions of the garden.
Solution:
Area = l×wl \times w:
w(w+10)=200.w(w + 10) = 200. w2+10w−200=0.w^2 + 10w – 200 = 0.
Factorize:
w2+20w−10w−200=0.w^2 + 20w – 10w – 200 = 0. w(w+20)−10(w+20)=0.w(w + 20) – 10(w + 20) = 0. (w−10)(w+20)=0.(w – 10)(w + 20) = 0. w=10 (valid), w=−20 (invalid).w = 10 \, \text{(valid)}, \, w = -20 \, \text{(invalid)}.
If w=10w = 10:
l=w+10=20 meters.l = w + 10 = 20 \, \text{meters}.
Answer: Width = 10 meters10 \, \text{meters}, Length = 20 meters20 \, \text{meters}.
5. Engineering: Structural Design
The stress S(x)S(x) on a beam is modeled by:
S(x)=−4×2+16x+20,S(x) = -4x^2 + 16x + 20,
where xx is the load in tons.
Problem 1: Maximum Stress
Find the maximum stress and the load at which it occurs.
Solution:
Maximum stress occurs at x=−b2ax = -\frac{b}{2a}, where a=−4a = -4, b=16b = 16:
x=−162(−4)=2 tons.x = -\frac{16}{2(-4)} = 2 \, \text{tons}.
Substitute x=2x = 2 into S(x)S(x):
S(2)=−4(2)2+16(2)+20.S(2) = -4(2)^2 + 16(2) + 20. S(2)=−16+32+20=36.S(2) = -16 + 32 + 20 = 36.
Answer: Maximum stress = 36 units36 \, \text{units}, occurs at 2 tons2 \, \text{tons}.
These examples showcase the versatility of polynomials in modeling real-world problems. Let me know if you’d like to work on additional examples or dive into another field of application!
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