Answer to Question #282207 in Calculus for john

Question

1. the sum of two numbers is 12. find the minimum value of the sum of their cubes.

2. a triangular corner lot has perpendicular sides of lengths 120m and 160m, resectively.

a. find the area of the longest rectangular building that can be constructed on the lot with sides parallel to the street.

b. find the perimeter that closes the building.

c. if it costs P6,000.00 per square meter of floor area of the building. how much would be the approximate cost if a three-storey building is to be constructes.

Answer 1

1. Let $x=$ the first number. Then the second number will be $12-x.$

The sum of their cubes will be

$$f(x)=x^3+(12-x)^3$$

$$=x^3+1728-432x+36x^2-x^3$$

$$=36(x^2-12x+48)$$

$$=36((x-6)^2+12)$$

$$=36(x-6)^2+432$$

This quadratic function has the absolute minimum with value of $432$ at $x=6.$

The minimum value of the sum of cubes of two given numbers is $432.$

2. Let $x=$ the width of the building, $y=$ the length of the building.

a. Right triangles $EBF$ and $ABC$ are similar

$$\dfrac{EF}{FB}=\dfrac{AC}{BC}$$

Substitute

$$\dfrac{y}{160-x}=\dfrac{120}{160}$$

$$y=\dfrac{3}{4}(160-x)$$

The area of the base of the building is

$$A=xy, 0<x< 160, 0<y<120$$

$$A=A(x)=\dfrac{3}{4}x(160-x)=-\dfrac{3}{4}((x-80)^2-6400)$$

$$=-\dfrac{3}{4}((x-80)^2-6400)$$

$$=-\dfrac{3}{4}(x-80)^2+4800$$

This quadratic function has the absolute maximum with value of $4800$ at $x=80.$

$$y=\dfrac{3}{4}(160-80)=60$$

$4800\ m^2$

b.

$$P=2(80+60)=280$$

$280\ m$

c.

$$3(4800)(6000)=86400000$$

$P86400000.00$