How would you go about solving this problem below?

Hi,

The height is 4.5 inches

First the rectangle with the largest area is always a square, so the base will be a square with sides of "x".

To find that, let x = length of a side of the base and h = height of the box. Then its volume is found by V = hx². This would become 729/2 = hx². Solving this for h gives h = 729/(2x²).

To express the total surface area, we want the area of the base plus 4 times the area of a side. This would be

Y = x² + 4(hx)

Now substitute 729/(2x²) for h.

Y = x² + 4(729/(2x²)* x)

This simplifies to

Y = x² + 1458/x

If you graph this and use the minimum command on the TI84, you will find the minimum surface area is when x = 9. To find h, use the formula 729/2 = hx² and substitute 9 for x.

729/2 = h9²

729/2 = h81

4.5 = h, so the height that gives the minimum surface area is 4.5 ft.

I hope that helps!! :-)

All the above problems involve 2 unknowns and one equation. So you cannot get a numerical solution. Best you could get is f(x) or f(y) meaning one variable in terms of the other one. However these are non linear equations so a close form solution is not guaranteed. Lets do the last one: log(y*y + x) = y ----> invert the log on both sides y * y + x = 10 ^ y x = 10 ^ y - y ^ 2 Another approach is to numerically graph y vs. x. At least gives you an idea what f(x) looks like.

V=x*y*z =729/2

S= xy +2(xz+yz)

Minimum S with restraint xyz-729/2=0

Using lagrangian multipliers

F(,x,y,z,k)= xy+2(xz+yz)+k(xyz-729/2)

Fx= y+2z+kyz=0

Fy= x+2z +kxz=0

Fz= 2(x+y) +k xy=0

yx+2zx+k*729/2=0

xy+2zy+k*729/2=0

2z(x-y)=0 so x=y and 4x+kx^2=0 so 4+kx=0 x=-4/k

z(2+kx)=-x and as kx=-4 2z=x and z=x/2

finally

x^3=729 and x =9 and z= 4.5 feet