# Thread: Word problems #2

1. ## Word problems #2

I am aware it is late so i will keep this short as possible:
1.According to postal regulations, the girth plus the length of a parcel sent by fourth-class mail may not exceed 108 in. What is the largest possible volume of a rectangular parcel with two square sides that can be sen by fourth-class mail? The answer says 11664in^3 = 6.75 ft^3.

2.A poster is to contain 108 cm^2 of printed matter with margins of 6cm each at top and bottom and 2 cm on the sides. What is the minimum cost of the of the poster if it is to be made of material costing 20 cents/cm^2? ans. is $60 2. 1. So $V = x^{2}y$ which we want to maximize. Then $4x+y \leq 108$. Note that girth is $2(h+w)$ or in this case $2(x+x) = 4x$. To get the maximum volume we set $4x+y = 108$. Then $y = 108 - 4x$ and $0 \leq x \leq 27$. Then $V = x^2(108-4x)$. $V'(x) = 216x-12x^2$. $0 = 216x-12x^2 = 12x(18-x)$. Critical values are $x = 0, \ x = 18$. Checking the endpoints $0,27$ (volume is 0) we conclude that the maximum volume occurs when $x = 18$. So $V_{\text{max}} = 11,644 \ \text{in}^3$. 2. So we want to minimize $C = 20(x-4)(y-12)$. Since $xy = 108$, $y = \frac{108}{x}$. Then $C(x) = 20(x-4)\left(\frac{108}{x}-12 \right)$. So $C'(x) = \frac{2160}{x} - 240 - \frac{2160(x-4)}{x^2}$. Critical values are $6,-6$. We get$60 when we plug in $C(-6)$.