Question

The water height of a pool is determined by 8g2 + 3g - 4, the rate that the pool is filled, and 9g2 - 2g - 5, the rate that water leaves the pool, where g represents the number of gallons entering or leaving the pool per minute.

Answer the folowing questions:

A. Write an expression that determines the height of the water in the pool.

B. What will be the height of the water if g = 1, 2, 3, and 4?

C. To the nearest tenth, at which value for g will the water reach its greatest height? explain.

Please i need help i need to turn this in for tomorrow!

Answer 1

`(a)H(g)=-(g^3)/(3)+ (5g^2)/(2)+g\\(b)H(1)=3.17\\H(2)=9.33\\H(3)=16.5\\H(4)=22.67\\(c)g=5.19 gallons`

Explanation:

(A)

Rate at which pool is filled
`= 8g\² + 3g - 4`

Rate at which water leaves the pool
`= 9g\² - 2g - 5`

The rate at which the height of the pool is changing =Rate In-Rate Out

`\alpha = 8g\² + 3g - 4 -( 9g\² - 2g - 5)\\=8g^2+3g-4-9g^2+2g+5\\(dH(g))/(dg) =-g^2+5g+1\\H(g)=\int(-g^2+5g+1)dg\\H(g)=-(g^3)/(3)+ (5g^2)/(2)+g`

(B)

Let water height be H

`H(g)=-(g^3)/(3)+ (5g^2)/(2)+g\\H(1)=-(1^3)/(3)+ (5^2)/(2)+1=3.17\\H(2)=-(2^3)/(3)+ (5(2)^2)/(2)+2=9.33\\H(3)=-(3^3)/(3)+ (5(3)^2)/(2)+3=16.5\\H(4)=-(4^3)/(3)+ (5(4)^2)/(2)+4=22.67`

(C)The water will reach its greatest height when the derivative of the Height Function is zero.

`(dH(g))/(dg) =-g^2+5g+1=0\\-g^2+5g+1=0\\g=-0.19,5.19\\`

At g=5.19 gallons the height of the water will reach its greatest height.

Answer 2

Part A

From the question, we know that

Rate at which pool is filled = 8g² + 3g - 4

Rate at which water leaves the pool = 9g² - 2g - 5

Hence, height of the pool would be = Rate at which pool is filled - Rate at which water leaves the pool

⇒ Height of the pool = 8g² + 3g - 4 - (9g² - 2g - 5)

⇒ Height of the pool = -g² + 5g + 1

Part B

Let water height be H

H (g) = -g² + 5g + 1

When g=1, H(1) = -(1)² + 5 (1) + 1 = 5 units

When g=2, H(2) = -(2)² + 5 (2) + 1 = -4+11 = 7 units

When g=3, H(3) = -(3)² + 5 (3) + 1 = -9+16 = 7 units

When g=4, H(4) = -(4)² + 5 (4) + 1 = -16+21 = 5 units

Part C

Now, we need to determine the value for g for which height would be maximum.

Looking at the expression that determines height, -g² + 5g + 1, we see that the coefficient of g² is negative. Hence the equation would represent a downward facing parabola, which means that the function H(g) will have a maxima point.

To find out the maxima point, differentiate H(g) with respect to g, and equate the resulting expression to zero.

`(d(H(g))/(dg)` = -2g + 5 = 0

⇒ 2g = 5

⇒ g = 2.5

So at g = 2.5, the height of the water in the pool is maximum. [Note: nearest tenth means rounding till the first decimal point, hence the answer is g=2.5]