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

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.

$\frac{d(H(g)}{dg}$ = -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]

Answer 2

Answer:

$(a)H(g)=-\frac{g^3}{3}+ \frac{5g^2}{2}+g\\(b)H(1)=3.17\\H(2)=9.33\\H(3)=16.5\\H(4)=22.67\\(c)g=5.19 gallons$

Step-by-step 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

$\alpha = 8g\² + 3g - 4 -( 9g\² - 2g - 5)\\=8g^2+3g-4-9g^2+2g+5\\\frac{dH(g)}{dg} =-g^2+5g+1\\H(g)=\int(-g^2+5g+1)dg\\H(g)=-\frac{g^3}{3}+ \frac{5g^2}{2}+g$

(B)

Let water height be H

$H(g)=-\frac{g^3}{3}+ \frac{5g^2}{2}+g\\H(1)=-\frac{1^3}{3}+ \frac{5^2}{2}+1=3.17\\H(2)=-\frac{2^3}{3}+ \frac{5(2)^2}{2}+2=9.33\\H(3)=-\frac{3^3}{3}+ \frac{5(3)^2}{2}+3=16.5\\H(4)=-\frac{4^3}{3}+ \frac{5(4)^2}{2}+4=22.67$

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

$\frac{dH(g)}{dg} =-g^2+5g+1=0\\-g^2+5g+1=0\\g=-0.19,5.19$

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