# Thread: solving optimization word problem?

1. ## solving optimization word problem?

A rectangular banner has a red border and a white center. The banner has an 8 inch border at the top and the bottom, and it has a 6 inch border on each side. The total area of the banner is 27 ft squared or 3888 in squared. What should be the dimensions of the banner (x and y) if the area of the white center is to be a maximum?

2. Originally Posted by aldricj

A rectangular banner has a red border and a white center. The banner has an 8 inch border at the top and the bottom, and it has a 6 inch border on each side. The total area of the banner is 27 ft squared or 3888 in squared. What should be the dimensions of the banner (x and y) if the area of the white center is to be a maximum?
let x = top length of white center
y = side length of white center

$(x+12)(y+16) = 3888$

$\displaystyle x+12 = \frac{3888}{y+16}$

$\displaystyle x = \frac{3888}{y+16} - 12$

maximize the area of the white center ...

$\displaystyle A = xy = \left(\frac{3888}{y+16} - 12\right)y$

take it from here?

3. okay, im pretty sure the next step is to take the derivative, but im not positive on how to do that. thanks for the response!

4. Originally Posted by aldricj
okay, im pretty sure the next step is to take the derivative, but im not positive on how to do that. thanks for the response!
$A = \dfrac{3888y}{y+16} - 12y$

quotient rule for the first (rational) term; derivative of the second term is -12, correct?

set $\dfrac{dA}{dy} = 0$ and determine the value of y that maximizes $A$

5. okay, i guess i should just say i need to see the step by step process to do this problem. im getting frustrated not being able to find the answer

6. $A = \dfrac{3888y}{y+16} - 12y$

$\dfrac{dA}{dy} = 3888 \cdot \dfrac{(y+16)(1)-(y)(1)}{(y+16)^2} - 12 = 0$

$\dfrac{16}{(y+16)^2} = \dfrac{12}{3888}
$

$\dfrac{16}{(y+16)^2} = \dfrac{1}{324}$

$(y+16)^2 = 16 \cdot 324
$

can you finish?